Unit 4 · Units · 12 min read
Trigonometry: Multiple, Submultiple and Transformation Formulas
What the syllabus asks for 5
- Compound angles: a quick review
- Multiple angles 2A and 3A
- Submultiple angles A/2
- Transformation formulas: sums, differences and products
- Exercises of the SEE type
In Grade 9 you learnt how to find \sin(A+B) and \cos(A-B) from the ratios of A and B. Those compound-angle formulas are the seed of this whole chapter. Put B=A and you get the ratios of the double angle 2A and the triple angle 3A (the multiple angles). Put A for \frac{A}{2} and you can go backwards, from A to the half angle \frac{A}{2} (a submultiple angle). Finally, adding or subtracting two sines or cosines can be turned into a product, and a product into a sum (the transformation formulas). With these tools you can find exact values such as \sin15^\circ and \tan22.5^\circ and prove many identities.
Typical questions: (i) find \sin2A,\ \cos2A,\ \tan2A from one given ratio; (ii) find the ratios of \theta when a ratio of \frac{\theta}{2} is given; (iii) prove an identity using 2A, 3A, \frac{A}{2} formulas; (iv) express a sum as a product or a product as a sum; (v) find the exact value of a product such as \cos20^\circ\cos40^\circ\cos80^\circ. In a proof, work on the left side only, show each step, and finish on the right side.
4.1 Compound angles: a quick review
For any angles A and B:
| \sin(A+B)=\sin A\cos B+\cos A\sin B | \sin(A-B)=\sin A\cos B-\cos A\sin B |
| \cos(A+B)=\cos A\cos B-\sin A\sin B | \cos(A-B)=\cos A\cos B+\sin A\sin B |
| \tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B} | \tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B} |
Memory aid: the sine formula keeps the sign of the angle and mixes (sin cos + cos sin); the cosine formula repeats (cos cos, sin sin) and gives the opposite sign.
Standard values:
| Angle | 0^\circ | 30^\circ | 45^\circ | 60^\circ | 90^\circ |
|---|---|---|---|---|---|
| \sin | 0 | \frac{1}{2} | \frac{1}{\sqrt{2}} | \frac{\sqrt{3}}{2} | 1 |
| \cos | 1 | \frac{\sqrt{3}}{2} | \frac{1}{\sqrt{2}} | \frac{1}{2} | 0 |
| \tan | 0 | \frac{1}{\sqrt{3}} | 1 | \sqrt{3} | not defined |
Problem. Find the exact values of \sin75^\circ and \tan105^\circ.
Solution. \sin75^\circ=\sin(45^\circ+30^\circ)=\sin45^\circ\cos30^\circ+\cos45^\circ\sin30^\circ
\tan105^\circ=\tan(60^\circ+45^\circ)=\frac{\sqrt{3}+1}{1-\sqrt{3}\cdot1}=\frac{\sqrt{3}+1}{1-\sqrt{3}}. Multiply top and bottom by (1+\sqrt{3}): \frac{(\sqrt{3}+1)^2}{1-3}=\frac{4+2\sqrt{3}}{-2}=-(2+\sqrt{3}).
Problem. If \sin A=\frac{3}{5} and \cos B=\frac{5}{13} with A and B acute, find \sin(A+B), \cos(A+B) and \sin(A-B).
Solution. Using \sin^2+\cos^2=1: \cos A=\sqrt{1-\frac{9}{25}}=\frac{4}{5} and \sin B=\sqrt{1-\frac{25}{169}}=\frac{12}{13}.
\sin(A+B)=\frac{3}{5}\cdot\frac{5}{13}+\frac{4}{5}\cdot\frac{12}{13}=\frac{15+48}{65}=\frac{63}{65}.
\cos(A+B)=\frac{4}{5}\cdot\frac{5}{13}-\frac{3}{5}\cdot\frac{12}{13}=\frac{20-36}{65}=-\frac{16}{65}.
\sin(A-B)=\frac{3}{5}\cdot\frac{5}{13}-\frac{4}{5}\cdot\frac{12}{13}=\frac{15-48}{65}=-\frac{33}{65}.
Problem. Prove that (i) \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B; (ii) if \tan A=\frac{1}{2} and \tan B=\frac{1}{3} then A+B=45^\circ.
Solution. (i) \sin(A+B)\sin(A-B)=(\sin A\cos B)^2-(\cos A\sin B)^2=\sin^2A(1-\sin^2B)-(1-\sin^2A)\sin^2B =\sin^2A-\sin^2A\sin^2B-\sin^2B+\sin^2A\sin^2B=\sin^2A-\sin^2B.
(ii) \tan(A+B)=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\cdot\frac{1}{3}}=\frac{\frac{5}{6}}{\frac{5}{6}}=1. Since A,B are acute, A+B=45^\circ.
Problem. Prove that (i) \cos(A+B)\cos(A-B)=\cos^2A-\sin^2B; (ii) \tan(45^\circ+A)=\frac{1+\tan A}{1-\tan A}. Hence find \tan15^\circ.
Solution. (i) \cos(A+B)\cos(A-B)=(\cos A\cos B)^2-(\sin A\sin B)^2=\cos^2A(1-\sin^2B)-(1-\cos^2A)\sin^2B=\cos^2A-\sin^2B.
(ii) \tan(45^\circ+A)=\frac{\tan45^\circ+\tan A}{1-\tan45^\circ\tan A}=\frac{1+\tan A}{1-\tan A}.
For \tan15^\circ=\tan(45^\circ-30^\circ)=\frac{1-\tan30^\circ}{1+\tan30^\circ}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{(\sqrt{3}-1)^2}{2}=2-\sqrt{3}.
4.2 Multiple angles 2A and 3A
A multiple angle of A is 2A,\ 3A,\ 4A,\ \ldots Putting B=A in the compound formulas gives the double-angle formulas.
We start from the compound-angle formulas with B=A.
Step 1 — sine.
Step 2 — cosine.
Step 3 — two more forms of \cos2A. Replace \sin^2A by 1-\cos^2A, or \cos^2A by 1-\sin^2A:
Step 4 — tangent.
Step 5 — in terms of t=\tan A. Divide 2\sin A\cos A by 1=\sin^2A+\cos^2A, then divide top and bottom by \cos^2A:
What it means. The double angle can be found from any one ratio of A. Which form of \cos2A to use depends on what is given: use 1-2\sin^2A if \sin A is known, 2\cos^2A-1 if \cos A is known.
Also: 1+\cos2A=2\cos^2A and 1-\cos2A=2\sin^2A.
Problem. If \sin A=\frac{3}{5} and A is acute, find \sin2A, \cos2A and \tan2A.
Solution. \cos A=\sqrt{1-\frac{9}{25}}=\frac{4}{5}.
\sin2A=2\sin A\cos A=2\cdot\frac{3}{5}\cdot\frac{4}{5}=\frac{24}{25}.
\cos2A=1-2\sin^2A=1-2\cdot\frac{9}{25}=\frac{7}{25}.
\tan2A=\frac{\sin2A}{\cos2A}=\frac{24}{7}.
Problem. (i) If \tan A=\frac{3}{4}, find \tan2A. (ii) If \tan\theta=\frac{1}{3}, find \cos2\theta.
Solution. (i) \tan2A=\frac{2\cdot\frac{3}{4}}{1-\frac{9}{16}}=\frac{\frac{3}{2}}{\frac{7}{16}}=\frac{3}{2}\cdot\frac{16}{7}=\frac{24}{7}.
(ii) \cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}=\frac{1-\frac{1}{9}}{1+\frac{1}{9}}=\frac{\frac{8}{9}}{\frac{10}{9}}=\frac{4}{5}.
Problem. Prove that (i) \frac{1-\cos2A}{\sin2A}=\tan A; (ii) (\sin A+\cos A)^2=1+\sin2A.
Solution. (i) LHS =\frac{2\sin^2A}{2\sin A\cos A}=\frac{\sin A}{\cos A}=\tan A= RHS.
(ii) LHS =\sin^2A+2\sin A\cos A+\cos^2A=1+\sin2A= RHS.
Problem. Prove that \frac{\cos A+\sin A}{\cos A-\sin A}-\frac{\cos A-\sin A}{\cos A+\sin A}=2\tan2A.
Solution. Put over a common denominator:
Problem. (i) If \cos A=-\frac{5}{13} and A lies in the second quadrant, find \sin2A and \cos2A. (ii) If \sin A+\cos A=\frac{1}{2}, find \sin2A. (iii) If \cos2A=\frac{7}{25} and A is acute, find \sin A.
Solution. (i) In the second quadrant \sin A>0: \sin A=\sqrt{1-\frac{25}{169}}=\frac{12}{13}. \sin2A=2\cdot\frac{12}{13}\cdot\left(-\frac{5}{13}\right)=-\frac{120}{169} and \cos2A=2\cos^2A-1=\frac{50}{169}-1=-\frac{119}{169}. Check: (-120)^2+(-119)^2=14400+14161=28561=169^2, so \sin^22A+\cos^22A=1 ✓.
(ii) Square both sides: \sin^2A+2\sin A\cos A+\cos^2A=\frac{1}{4}, so 1+\sin2A=\frac{1}{4} and \sin2A=-\frac{3}{4}.
(iii) 1-2\sin^2A=\frac{7}{25}\Rightarrow\sin^2A=\frac{9}{25}, so \sin A=\frac{3}{5} (acute).
Problem. Prove that (i) \cos^4A-\sin^4A=\cos2A; (ii) \tan A+\cot A=\frac{2}{\sin2A}; (iii) \frac{\sin2A}{1-\cos2A}=\cot A.
Solution. (i) \cos^4A-\sin^4A=(\cos^2A-\sin^2A)(\cos^2A+\sin^2A)=\cos2A\cdot1=\cos2A.
(ii) \tan A+\cot A=\frac{\sin^2A+\cos^2A}{\sin A\cos A}=\frac{1}{\sin A\cos A}=\frac{2}{2\sin A\cos A}=\frac{2}{\sin2A}.
(iii) \frac{2\sin A\cos A}{2\sin^2A}=\frac{\cos A}{\sin A}=\cot A.
Triple angle
We start from 3A=2A+A and use the compound and double-angle formulas.
Step 1 — sine.
Step 2 — write \cos^2A as 1-\sin^2A.
Step 3 — collect terms.
Step 4 — cosine.
Step 5 — write \sin^2A as 1-\cos^2A.
What it means. The triple angle needs only the single ratio of A. Check with A=30^\circ: 3\sin30^\circ-4\sin^330^\circ=\frac{3}{2}-\frac{1}{2}=1=\sin90^\circ. ✓
Problem. (i) If \cos A=\frac{1}{2}, find \cos3A. (ii) If \tan A=\frac{1}{2}, find \tan3A. (iii) Prove that \sin3A=4\sin A\sin(60^\circ-A)\sin(60^\circ+A).
Solution. (i) \cos3A=4\left(\frac{1}{2}\right)^3-3\cdot\frac{1}{2}=\frac{1}{2}-\frac{3}{2}=-1. (Here A=60^\circ and \cos180^\circ=-1. ✓)
(ii) \tan3A=\frac{3\cdot\frac{1}{2}-\frac{1}{8}}{1-3\cdot\frac{1}{4}}=\frac{\frac{11}{8}}{\frac{1}{4}}=\frac{11}{2}.
(iii) First, \sin(60^\circ-A)\sin(60^\circ+A)=\sin^260^\circ-\sin^2A=\frac{3}{4}-\sin^2A (by the identity proved in 4.1). So
- Writing \sin2A=2\sin A or \cos2A=2\cos A. Doubling the angle does not double the ratio. Test with A=30^\circ: \sin60^\circ=0.866\neq2\sin30^\circ=1.
- Writing \cos2A=\cos^2A+\sin^2A. That equals 1. The correct form has a minus sign.
- Using \sin3A=3\sin A-4\sin A^3 or forgetting that \sin^3A means (\sin A)^3.
- Choosing the wrong form of \cos2A and getting a long calculation: match the form to the given ratio.
4.3 Submultiple angles \frac{A}{2}
A submultiple angle of A is \frac{A}{2},\ \frac{A}{3},\ \ldots Replace A by \frac{A}{2} in the double-angle formulas.
We start from the double-angle formulas written for the angle \frac{A}{2} (so 2\times\frac{A}{2}=A).
Step 1 — sine.
Step 2 — cosine, two forms.
Step 3 — solve for the squares.
Step 4 — divide.
Step 5 — a form without a square root. Multiply top and bottom of \tan\frac{A}{2}=\frac{\sin\frac{A}{2}}{\cos\frac{A}{2}} by 2\cos\frac{A}{2} (or by 2\sin\frac{A}{2}):
What it means. The sign of the root (\pm) is decided by the quadrant of \frac{A}{2}, not of A. For acute \frac{A}{2} all three ratios are positive.
For the third part: \sin A=3\sin\frac{A}{3}-4\sin^3\frac{A}{3} and \cos A=4\cos^3\frac{A}{3}-3\cos\frac{A}{3}.
Problem. (i) If \cos\frac{\theta}{2}=\frac{4}{5} (\frac{\theta}{2} acute), find \sin\theta, \cos\theta, \tan\theta. (ii) If \tan\frac{\theta}{2}=\frac{4}{3}, find \sin\theta, \cos\theta. (iii) If \sin\frac{\theta}{2}=\frac{1}{\sqrt{2}}, find \cos\theta.
Solution. (i) \sin\frac{\theta}{2}=\sqrt{1-\frac{16}{25}}=\frac{3}{5}. \sin\theta=2\cdot\frac{3}{5}\cdot\frac{4}{5}=\frac{24}{25}; \cos\theta=2\cos^2\frac{\theta}{2}-1=\frac{32}{25}-1=\frac{7}{25}; \tan\theta=\frac{24}{7}.
(ii) With t=\frac{4}{3}: 1+t^2=\frac{25}{9}. \sin\theta=\frac{2t}{1+t^2}=\frac{\frac{8}{3}}{\frac{25}{9}}=\frac{24}{25}. \cos\theta=\frac{1-t^2}{1+t^2}=\frac{-\frac{7}{9}}{\frac{25}{9}}=-\frac{7}{25}. (So \theta is in the second quadrant; \tan\theta=-\frac{24}{7}.)
(iii) \cos\theta=1-2\sin^2\frac{\theta}{2}=1-2\cdot\frac{1}{2}=0. (Here \frac{\theta}{2}=45^\circ, \theta=90^\circ. ✓)
Problem. Find the exact values of \sin15^\circ and \tan22.5^\circ.
Solution. \sin^215^\circ=\frac{1-\cos30^\circ}{2}=\frac{1-\frac{\sqrt{3}}{2}}{2}=\frac{2-\sqrt{3}}{4}, so \sin15^\circ=\frac{\sqrt{2-\sqrt{3}}}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}. (This matches the compound-angle value from \sin(45^\circ-30^\circ).)
\tan22.5^\circ=\frac{\sin45^\circ}{1+\cos45^\circ}=\frac{\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}}=\frac{1}{\sqrt{2}+1}=\sqrt{2}-1.
Problem. Prove that (i) \frac{1-\cos\theta}{1+\cos\theta}=\tan^2\frac{\theta}{2}; (ii) \frac{\sin\theta}{1+\cos\theta}=\tan\frac{\theta}{2}; (iii) \sqrt{\frac{1-\cos\theta}{2}}\cdot\sqrt{\frac{1+\cos\theta}{2}}=\frac{\sin\theta}{2} (for 0<\theta<180^\circ).
Solution. (i) LHS =\frac{2\sin^2\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}}=\tan^2\frac{\theta}{2}.
(ii) LHS =\frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}}=\tan\frac{\theta}{2}.
(iii) LHS =\sqrt{\frac{1-\cos^2\theta}{4}}=\frac{\sin\theta}{2}.
Problem. (i) If \cos\theta=\frac{7}{25} (\theta acute), find \sin\frac{\theta}{2} and \cos\frac{\theta}{2}. (ii) Prove that \tan\left(45^\circ+\frac{\theta}{2}\right)=\sec\theta+\tan\theta. (iii) Prove that \frac{\sin\theta}{1-\cos\theta}=\cot\frac{\theta}{2}.
Solution. (i) \sin^2\frac{\theta}{2}=\frac{1-\frac{7}{25}}{2}=\frac{9}{25}, so \sin\frac{\theta}{2}=\frac{3}{5}. \cos^2\frac{\theta}{2}=\frac{1+\frac{7}{25}}{2}=\frac{16}{25}, so \cos\frac{\theta}{2}=\frac{4}{5} (both positive since \frac{\theta}{2} is acute).
(ii) Let t=\tan\frac{\theta}{2}. Then \tan\left(45^\circ+\frac{\theta}{2}\right)=\frac{1+t}{1-t}=\frac{(1+t)^2}{(1-t)(1+t)}=\frac{1+t^2}{1-t^2}+\frac{2t}{1-t^2}=\sec\theta+\tan\theta, using \cos\theta=\frac{1-t^2}{1+t^2} and \tan\theta=\frac{2t}{1-t^2}.
(iii) \frac{\sin\theta}{1-\cos\theta}=\frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}}=\frac{\cos\frac{\theta}{2}}{\sin\frac{\theta}{2}}=\cot\frac{\theta}{2}.
4.4 Transformation formulas: sums, differences and products
We start from the four compound-angle formulas for sine and cosine.
Step 1 — add the two sine formulas.
Step 2 — subtract them.
Step 3 — add the two cosine formulas.
Step 4 — subtract them (second minus first).
Step 5 — sum to product. Put A+B=C and A-B=D, so A=\frac{C+D}{2} and B=\frac{C-D}{2}. Steps 1 to 4 become the formulas below.
What it means. A product of two ratios turns into a sum or difference of single ratios (easier to add or integrate), and a sum or difference turns into a product (easier to factorise and simplify).
Product to sum
Sum to product
(1) writing which of the four product-to-sum formulas you start from, (2) choosing C and D so that \frac{C+D}{2} and \frac{C-D}{2} are nice angles, (3) the minus sign in \cos C-\cos D, (4) a simplified final value.
Problem. Find the exact values of (i) \sin75^\circ+\sin15^\circ, (ii) \cos15^\circ-\cos75^\circ.
Solution. (i) \sin75^\circ+\sin15^\circ=2\sin\frac{75^\circ+15^\circ}{2}\cos\frac{75^\circ-15^\circ}{2}=2\sin45^\circ\cos30^\circ=2\cdot\frac{1}{\sqrt{2}}\cdot\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{\sqrt{2}}=\frac{\sqrt{6}}{2}.
(ii) \cos15^\circ-\cos75^\circ=-2\sin\frac{15^\circ+75^\circ}{2}\sin\frac{15^\circ-75^\circ}{2}=-2\sin45^\circ\sin(-30^\circ)=2\cdot\frac{1}{\sqrt{2}}\cdot\frac{1}{2}=\frac{1}{\sqrt{2}}.
Problem. Express as a sum or difference: (i) 2\sin5\theta\cos2\theta, (ii) 2\cos7\theta\cos3\theta, (iii) \sin75^\circ\sin15^\circ (and evaluate).
Solution. (i) 2\sin5\theta\cos2\theta=\sin7\theta+\sin3\theta.
(ii) 2\cos7\theta\cos3\theta=\cos10\theta+\cos4\theta.
(iii) \sin75^\circ\sin15^\circ=\frac{1}{2}\left[\cos60^\circ-\cos90^\circ\right]=\frac{1}{2}\left[\frac{1}{2}-0\right]=\frac{1}{4}.
Problem. Prove that \frac{\sin5A+\sin3A}{\cos5A+\cos3A}=\tan4A.
Solution. \sin5A+\sin3A=2\sin4A\cos A and \cos5A+\cos3A=2\cos4A\cos A. So
Problem. Show that (i) \cos20^\circ\cos40^\circ\cos80^\circ=\frac{1}{8}; (ii) \sin20^\circ\sin40^\circ\sin80^\circ=\frac{\sqrt{3}}{8}.
Solution. (i) Let P=\cos20^\circ\cos40^\circ\cos80^\circ. Multiply and divide by 2\sin20^\circ and use 2\sin\theta\cos\theta=\sin2\theta repeatedly:
Since \sin160^\circ=\sin20^\circ, P=\frac{1}{8}.
(ii) Use \sin3A=4\sin A\sin(60^\circ-A)\sin(60^\circ+A) from 4.2 with A=20^\circ: \sin60^\circ=4\sin20^\circ\sin40^\circ\sin80^\circ. So the product is \frac{\sin60^\circ}{4}=\frac{\sqrt{3}}{8}.
Problem. Prove that (i) \frac{\sin3A-\sin A}{\cos3A+\cos A}=\tan A; (ii) \frac{\cos A-\cos3A}{\sin3A-\sin A}=\tan2A; (iii) \cos105^\circ+\cos15^\circ=\frac{1}{\sqrt{2}}.
Solution. (i) \sin3A-\sin A=2\cos2A\sin A and \cos3A+\cos A=2\cos2A\cos A. The ratio is \frac{\sin A}{\cos A}=\tan A.
(ii) \cos A-\cos3A=-2\sin2A\sin(-A)=2\sin2A\sin A and \sin3A-\sin A=2\cos2A\sin A. The ratio is \frac{\sin2A}{\cos2A}=\tan2A.
(iii) \cos105^\circ+\cos15^\circ=2\cos60^\circ\cos45^\circ=2\cdot\frac{1}{2}\cdot\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}.
4.5 Exercises of the SEE type
Problem. Prove that \sin A+\sin3A+\sin5A=\sin3A\,(1+2\cos2A).
Solution. \sin A+\sin5A=2\sin3A\cos2A. So LHS =2\sin3A\cos2A+\sin3A=\sin3A(2\cos2A+1) =\sin3A\,(1+2\cos2A) = RHS.
Problem. Prove that \cos4A=8\cos^4A-8\cos^2A+1.
Solution. \cos4A=\cos2(2A)=2\cos^22A-1. Now \cos2A=2\cos^2A-1, so
Problem. Prove that \cos A+\cos(120^\circ-A)+\cos(120^\circ+A)=0.
Solution. \cos(120^\circ-A)+\cos(120^\circ+A)=2\cos120^\circ\cos A=2\left(-\frac{1}{2}\right)\cos A=-\cos A. So the sum is \cos A-\cos A=0.
Problem. Find the value of \sin10^\circ\sin50^\circ\sin70^\circ.
Solution. Write 50^\circ=60^\circ-10^\circ and 70^\circ=60^\circ+10^\circ. Then \sin10^\circ\sin(60^\circ-10^\circ)\sin(60^\circ+10^\circ)=\frac{\sin30^\circ}{4}=\frac{1/2}{4}=\frac{1}{8}, by \sin3A=4\sin A\sin(60^\circ-A)\sin(60^\circ+A) with A=10^\circ.
Chapter summary
- Compound angles: \sin(A\pm B)=\sin A\cos B\pm\cos A\sin B; \cos(A\pm B)=\cos A\cos B\mp\sin A\sin B; \tan(A\pm B)=\frac{\tan A\pm\tan B}{1\mp\tan A\tan B}.
- Double angle: \sin2A=2\sin A\cos A=\frac{2\tan A}{1+\tan^2A}; \cos2A=\cos^2A-\sin^2A=2\cos^2A-1=1-2\sin^2A=\frac{1-\tan^2A}{1+\tan^2A}; \tan2A=\frac{2\tan A}{1-\tan^2A}.
- Triple angle: \sin3A=3\sin A-4\sin^3A; \cos3A=4\cos^3A-3\cos A; \tan3A=\frac{3\tan A-\tan^3A}{1-3\tan^2A}.
- Half angle: \sin^2\frac{A}{2}=\frac{1-\cos A}{2}; \cos^2\frac{A}{2}=\frac{1+\cos A}{2}; \tan\frac{A}{2}=\frac{\sin A}{1+\cos A}=\frac{1-\cos A}{\sin A}. Sign from the quadrant of \frac{A}{2}.
- Product to sum: 2\sin A\cos B=\sin(A+B)+\sin(A-B); 2\cos A\cos B=\cos(A+B)+\cos(A-B); 2\sin A\sin B=\cos(A-B)-\cos(A+B).
- Sum to product: \sin C+\sin D=2\sin\frac{C+D}{2}\cos\frac{C-D}{2}; \cos C+\cos D=2\cos\frac{C+D}{2}\cos\frac{C-D}{2}; \cos C-\cos D=-2\sin\frac{C+D}{2}\sin\frac{C-D}{2}.
- Special products: \cos20^\circ\cos40^\circ\cos80^\circ=\frac{1}{8}; \sin20^\circ\sin40^\circ\sin80^\circ=\frac{\sqrt{3}}{8}.
Practice questions
Group A — Multiple choice (1 mark each)
- The value of \sin75^\circ is [1]a\frac{\sqrt{6}-\sqrt{2}}{4}b\frac{\sqrt{6}+\sqrt{2}}{4}c\frac{\sqrt{3}+1}{2}d\frac{\sqrt{3}}{2}
- If \sin A=\frac{3}{5}, then \cos2A is [1]a\frac{24}{25}b\frac{16}{25}c\frac{7}{25}d\frac{9}{25}
- Which of these is not equal to \cos2A? [1]a\cos^2A-\sin^2Ab2\cos^2A-1c1-2\sin^2Ad2\sin^2A-1
- \sin^2\frac{A}{2} equals [1]a\frac{1+\cos A}{2}b\frac{1-\cos A}{2}c\frac{1-\sin A}{2}d1-\cos A
- 2\sin A\cos B equals [1]a\sin(A+B)+\sin(A-B)b\sin(A+B)-\sin(A-B)c\cos(A+B)+\cos(A-B)d\cos(A-B)-\cos(A+B)
- The value of \tan22.5^\circ is [1]a\sqrt{2}+1b\sqrt{2}-1c\frac{1}{\sqrt{2}}d\frac{\sqrt{3}}{3}
Answers to Group A
1. (b) — \sin(45^\circ+30^\circ)=\frac{\sqrt{6}+\sqrt{2}}{4}.
2. (c) — 1-2\cdot\frac{9}{25}=\frac{7}{25}.
3. (d) — 1-2\sin^2A is the correct form; 2\sin^2A-1=-\cos2A.
4. (b) — from \cos A=1-2\sin^2\frac{A}{2}.
5. (a) — add the two sine compound formulas.
6. (b) — \tan22.5^\circ=\frac{\sin45^\circ}{1+\cos45^\circ}=\sqrt{2}-1.
Group B — Short answer (2 marks each)
- Find the exact value of \sin15^\circ using a compound-angle formula. [2]
- If \tan A=\frac{3}{4}, find \tan2A. [2]
- Prove that \frac{\sin2A}{1+\cos2A}=\tan A. [2]
- Express 2\sin5\theta\cos2\theta as a sum. [2]
- Find the value of \sin75^\circ+\sin15^\circ. [2]
Answers to Group B
1. \sin(45^\circ-30^\circ)=\frac{1}{\sqrt{2}}\cdot\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\cdot\frac{1}{2}=\frac{\sqrt{3}-1}{2\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{4}.
2. \tan2A=\frac{2\cdot\frac{3}{4}}{1-\frac{9}{16}}=\frac{24}{7}.
3. LHS =\frac{2\sin A\cos A}{2\cos^2A}=\tan A.
4. 2\sin5\theta\cos2\theta=\sin7\theta+\sin3\theta.
5. 2\sin45^\circ\cos30^\circ=2\cdot\frac{1}{\sqrt{2}}\cdot\frac{\sqrt{3}}{2}=\frac{\sqrt{6}}{2}.
Group C — Short answer (3 marks each)
- If \sin A=\frac{3}{5} and A is acute, find \sin2A, \cos2A and \tan2A. [3]
- If \cos\frac{\theta}{2}=\frac{4}{5} (\frac{\theta}{2} acute), find \sin\theta and \cos\theta. [3]
- Prove that \frac{\sin5A+\sin3A}{\cos5A+\cos3A}=\tan4A. [3]
- Prove that \sin3A=4\sin A\sin(60^\circ-A)\sin(60^\circ+A). [3]
- Find the value of \cos15^\circ-\cos75^\circ. [3]
Answers to Group C
1. \cos A=\frac{4}{5}. \sin2A=2\cdot\frac{3}{5}\cdot\frac{4}{5}=\frac{24}{25}; \cos2A=1-2\cdot\frac{9}{25}=\frac{7}{25}; \tan2A=\frac{24}{7}.
2. \sin\frac{\theta}{2}=\frac{3}{5}. \sin\theta=2\cdot\frac{3}{5}\cdot\frac{4}{5}=\frac{24}{25}; \cos\theta=2\cdot\frac{16}{25}-1=\frac{7}{25}.
3. \sin5A+\sin3A=2\sin4A\cos A; \cos5A+\cos3A=2\cos4A\cos A; the ratio is \tan4A.
4. \sin(60^\circ-A)\sin(60^\circ+A)=\sin^260^\circ-\sin^2A=\frac{3}{4}-\sin^2A. Then 4\sin A(\frac{3}{4}-\sin^2A)=3\sin A-4\sin^3A=\sin3A.
5. -2\sin45^\circ\sin(-30^\circ)=2\cdot\frac{1}{\sqrt{2}}\cdot\frac{1}{2}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}.
Group D — Long answer (4 marks each)
- Prove that \frac{\cos A+\sin A}{\cos A-\sin A}-\frac{\cos A-\sin A}{\cos A+\sin A}=2\tan2A. [4]
- Prove that \cos20^\circ\cos40^\circ\cos80^\circ=\frac{1}{8}. [4]
- If \tan A=\frac{1}{2} and \tan B=\frac{1}{3}, show that A+B=45^\circ and find \tan2A. [4]
- If \tan\frac{\theta}{2}=\frac{4}{3}, find \sin\theta, \cos\theta and \tan\theta. [4]
- Prove that \sin20^\circ\sin40^\circ\sin80^\circ=\frac{\sqrt{3}}{8}. [4]
Answers to Group D
1. LHS =\frac{(\cos A+\sin A)^2-(\cos A-\sin A)^2}{\cos^2A-\sin^2A}=\frac{4\sin A\cos A}{\cos2A}=\frac{2\sin2A}{\cos2A}=2\tan2A.
2. P=\frac{2\sin20^\circ\cos20^\circ\cos40^\circ\cos80^\circ}{2\sin20^\circ}=\frac{\sin40^\circ\cos40^\circ\cos80^\circ}{2\sin20^\circ}=\frac{\sin80^\circ\cos80^\circ}{4\sin20^\circ}=\frac{\sin160^\circ}{8\sin20^\circ}=\frac{1}{8}.
3. \tan(A+B)=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}=1, so A+B=45^\circ. \tan2A=\frac{1}{1-\frac{1}{4}}=\frac{4}{3}.
4. t=\frac{4}{3}, 1+t^2=\frac{25}{9}. \sin\theta=\frac{8/3}{25/9}=\frac{24}{25}; \cos\theta=\frac{1-16/9}{25/9}=-\frac{7}{25}; \tan\theta=-\frac{24}{7}.
5. From \sin3A=4\sin A\sin(60^\circ-A)\sin(60^\circ+A) with A=20^\circ: \sin60^\circ=4\sin20^\circ\sin40^\circ\sin80^\circ, so the product is \frac{\sqrt{3}/2}{4}=\frac{\sqrt{3}}{8}.
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