Question

use the sum or difference identity to find an exact value.

1. tan15°
2. cos75°
3. sin(7π/12)
4. tan(11π/12)

write the expression as the sine, cosine or tangent of an angle.

1. cos94°cos18° + sin94°sin18°
2. sin42°cos17° - cos42°sin17°
3. (tan19° + tan43°)/(1 - tan19°tan43°)
4. sin(π/3)cos(π/7) + cos(π/3)sin(π/7)

Explanation:

Step1: Recall the sum - difference identities

The sum and difference identities for tangent, cosine and sine are:
- $\tan(A\pm B)=\frac{\tan A\pm\tan B}{1\mp\tan A\tan B}$
- $\cos(A - B)=\cos A\cos B+\sin A\sin B$
- $\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B$

Step2: Solve for $\tan15^{\circ}$

We know that $15^{\circ}=45^{\circ}- 30^{\circ}$.
Using the tangent difference formula $\tan(A - B)=\frac{\tan A-\tan B}{1 + \tan A\tan B}$, with $A = 45^{\circ}$ and $B=30^{\circ}$, $\tan45^{\circ}=1$ and $\tan30^{\circ}=\frac{\sqrt{3}}{3}$.
$\tan15^{\circ}=\tan(45^{\circ}-30^{\circ})=\frac{\tan45^{\circ}-\tan30^{\circ}}{1+\tan45^{\circ}\tan30^{\circ}}=\frac{1-\frac{\sqrt{3}}{3}}{1 + 1\times\frac{\sqrt{3}}{3}}=\frac{3 - \sqrt{3}}{3+\sqrt{3}}=\frac{(3 - \sqrt{3})(3-\sqrt{3})}{(3+\sqrt{3})(3 - \sqrt{3})}=\frac{9-6\sqrt{3}+3}{9 - 3}=\frac{12-6\sqrt{3}}{6}=2-\sqrt{3}$

Step3: Solve for $\cos75^{\circ}$

We know that $75^{\circ}=45^{\circ}+30^{\circ}$.
Using the cosine sum formula $\cos(A + B)=\cos A\cos B-\sin A\sin B$, with $A = 45^{\circ}$ and $B = 30^{\circ}$, $\cos45^{\circ}=\frac{\sqrt{2}}{2}$, $\sin45^{\circ}=\frac{\sqrt{2}}{2}$, $\cos30^{\circ}=\frac{\sqrt{3}}{2}$, $\sin30^{\circ}=\frac{1}{2}$.
$\cos75^{\circ}=\cos(45^{\circ}+30^{\circ})=\cos45^{\circ}\cos30^{\circ}-\sin45^{\circ}\sin30^{\circ}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}$

Step4: Solve for $\sin\frac{7\pi}{12}$

We know that $\frac{7\pi}{12}=\frac{\pi}{3}+\frac{\pi}{4}$.
Using the sine sum formula $\sin(A + B)=\sin A\cos B+\cos A\sin B$, with $A=\frac{\pi}{3}$ and $B = \frac{\pi}{4}$, $\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}$, $\cos\frac{\pi}{3}=\frac{1}{2}$, $\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}$, $\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}$.
$\sin\frac{7\pi}{12}=\sin(\frac{\pi}{3}+\frac{\pi}{4})=\sin\frac{\pi}{3}\cos\frac{\pi}{4}+\cos\frac{\pi}{3}\sin\frac{\pi}{4}=\frac{\sqrt{3}}{2}\times\frac{\sqrt{2}}{2}+\frac{1}{2}\times\frac{\sqrt{2}}{2}=\frac{\sqrt{6}+\sqrt{2}}{4}$

Step5: Solve for $\tan\frac{11\pi}{12}$

We know that $\frac{11\pi}{12}=\frac{3\pi}{4}+\frac{\pi}{6}$.
Using the tangent sum formula $\tan(A + B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$, with $\tan\frac{3\pi}{4}=- 1$ and $\tan\frac{\pi}{6}=\frac{\sqrt{3}}{3}$.
$\tan\frac{11\pi}{12}=\tan(\frac{3\pi}{4}+\frac{\pi}{6})=\frac{\tan\frac{3\pi}{4}+\tan\frac{\pi}{6}}{1-\tan\frac{3\pi}{4}\tan\frac{\pi}{6}}=\frac{-1+\frac{\sqrt{3}}{3}}{1-(-1)\times\frac{\sqrt{3}}{3}}=\frac{-3 + \sqrt{3}}{3+\sqrt{3}}=\frac{(-3+\sqrt{3})(3 - \sqrt{3})}{(3+\sqrt{3})(3 - \sqrt{3})}=\frac{-9 + 6\sqrt{3}-3}{9 - 3}=\frac{-12 + 6\sqrt{3}}{6}=\sqrt{3}-2$

Step6: Simplify $\cos94^{\circ}\cos18^{\circ}+\sin94^{\circ}\sin18^{\circ}$

Using the cosine difference formula $\cos(A - B)=\cos A\cos B+\sin A\sin B$, here $A = 94^{\circ}$ and $B=18^{\circ}$, so $\cos94^{\circ}\cos18^{\circ}+\sin94^{\circ}\sin18^{\circ}=\cos(94^{\circ}-18^{\circ})=\cos76^{\circ}$

Step7: Simplify $\sin42^{\circ}\cos17^{\circ}-\cos42^{\circ}\sin17^{\circ}$

Using the sine difference formula $\sin(A - B)=\sin A\cos B-\cos A\sin B$, here $A = 42^{\circ}$ and $B = 17^{\circ}$, so $\sin42^{\circ}\cos17^{\circ}-\cos42^{\circ}\sin17^{\circ}=\sin(42^{\circ}-17^{\circ})=\sin25^{\circ}$

Step8: Simplify $\frac{\tan19^{\circ}+\tan43^{\circ}}{1-\tan19^{\circ}\tan43^{\circ}}$

Using the tangent sum formula $\tan(A + B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$, here $A = 19^{\circ}$ and $B = 43^{\circ}$, so $\frac{\tan19^{\circ}+\tan43^{\circ}}{1-\tan19^{\circ}\tan43^{\circ}}=\tan(19^{\circ}+43^{\circ})=\tan62^{\circ}$

Step9: Simplify $\sin\frac{\pi}{3}\cos\frac{\pi}{7}+\cos\frac{\pi}{3}\sin\frac{\pi}{7}$

Using the sine sum formula $\sin(A + B)=\sin A\cos B+\cos A\sin B$, here $A=\frac{\pi}{3}$ and $B=\frac{\pi}{7}$, so $\sin\frac{\pi}{3}\cos\frac{\pi}{7}+\cos\frac{\pi}{3}\sin\frac{\pi}{7}=\sin(\frac{\pi}{3}+\frac{\pi}{7})=\sin\frac{7\pi + 3\pi}{21}=\sin\frac{10\pi}{21}$

Answer:

1. $\tan15^{\circ}=2 - \sqrt{3}$
2. $\cos75^{\circ}=\frac{\sqrt{6}-\sqrt{2}}{4}$
3. $\sin\frac{7\pi}{12}=\frac{\sqrt{6}+\sqrt{2}}{4}$
4. $\tan\frac{11\pi}{12}=\sqrt{3}-2$
5. $\cos94^{\circ}\cos18^{\circ}+\sin94^{\circ}\sin18^{\circ}=\cos76^{\circ}$
6. $\sin42^{\circ}\cos17^{\circ}-\cos42^{\circ}\sin17^{\circ}=\sin25^{\circ}$
7. $\frac{\tan19^{\circ}+\tan43^{\circ}}{1-\tan19^{\circ}\tan43^{\circ}}=\tan62^{\circ}$
8. $\sin\frac{\pi}{3}\cos\frac{\pi}{7}+\cos\frac{\pi}{3}\sin\frac{\pi}{7}=\sin\frac{10\pi}{21}$