9.2 Sum and difference identities  (Page 6/6)

Explain the basis for the cofunction identities and when they apply.

Is there only one way to evaluate $\cos \left(\frac{5\pi }{4}\right)?$ Explain how to set up the solution in two different ways, and then compute to make sure they give the same answer.

Explain to someone who has forgotten the even-odd properties of sinusoidal functions how the addition and subtraction formulas can determine this characteristic for $f(x)=\sin (x)$ and $g(x)=\cos (x).$ (Hint: $0-x=-x$ )

$\cos \left(\frac{7\pi }{12}\right)$

$\cos \left(\frac{\pi }{12}\right)$

$\sin \left(\frac{5\pi }{12}\right)$

$\sin \left(\frac{11\pi }{12}\right)$

$\tan \left(-\frac{\pi }{12}\right)$

$\tan \left(\frac{19\pi }{12}\right)$

$\sin \left(x+\frac{11\pi }{6}\right)$

$\sin \left(x-\frac{3\pi }{4}\right)$

$\cos \left(x-\frac{5\pi }{6}\right)$

$\cos \left(x+\frac{2\pi }{3}\right)$

$\csc \left(\frac{\pi }{2}-t\right)$

$\sec \left(\frac{\pi }{2}-\theta \right)$

$\cot \left(\frac{\pi }{2}-x\right)$

$\tan \left(\frac{\pi }{2}-x\right)$

$\sin (2x)\cos (5x)-\sin (5x)\cos (2x)$

$\frac{\tan \left(\frac{3}{2}x\right)-\tan \left(\frac{7}{5}x\right)}{1+\tan \left(\frac{3}{2}x\right)\tan \left(\frac{7}{5}x\right)}$

Given that $\sin a=\frac{2}{3}$ and $\cos b=-\frac{1}{4},$ with $a$ and $b$ both in the interval $\left[\frac{\pi }{2},\pi \right),$ find $\sin (a+b)$ and $\cos (a-b).$

Given that $\sin a=\frac{4}{5},$ and $\cos b=\frac{1}{3},$ with $a$ and $b$ both in the interval $\left[0,\frac{\pi }{2}\right),$ find $\sin (a-b)$ and $\cos (a+b).$

$\sin \left({\cos }^{-1}(0)-{\cos }^{-1}\left(\frac{1}{2}\right)\right)$

$\cos \left({\cos }^{-1}\left(\frac{\sqrt{2}}{2}\right)+{\sin }^{-1}\left(\frac{\sqrt{3}}{2}\right)\right)$

$\tan \left({\sin }^{-1}\left(\frac{1}{2}\right)-{\cos }^{-1}\left(\frac{1}{2}\right)\right)$

$\cos \left(\frac{\pi }{2}-x\right)$

$\sin (\pi -x)$

$\tan \left(\frac{\pi }{3}+x\right)$

$\sin \left(\frac{\pi }{3}+x\right)$

$\tan \left(\frac{\pi }{4}-x\right)$

$\cos \left(\frac{7\pi }{6}+x\right)$

$\sin \left(\frac{\pi }{4}+x\right)$

$\cos \left(\frac{5\pi }{4}+x\right)$

$f\left(x\right)=\sin \left(4x\right)-\sin \left(3x\right)\cos x,g\left(x\right)=\sin x\cos \left(3x\right)$

$f\left(x\right)=\cos \left(4x\right)+\sin x\sin \left(3x\right),g\left(x\right)=-\cos x\cos \left(3x\right)$

$f\left(x\right)=\sin \left(3x\right)\cos \left(6x\right),g\left(x\right)=-\sin \left(3x\right)\cos \left(6x\right)$

$f(x)=\sin (4x),g(x)=\sin (5x)\cos x-\cos (5x)\sin x$

$f(x)=\sin (2x),g(x)=2\sin x\cos x$

$f\left(\theta \right)=\cos \left(2\theta \right),g\left(\theta \right)={\cos }^2\theta -{\sin }^2\theta$

$f(\theta )=\tan (2\theta ),g(\theta )=\frac{\tan \theta }{1+{\tan }^2\theta }$

$f(x)=\sin (3x)\sin x,g(x)={\sin }^2(2x){\cos }^{2}x-{\cos }^2(2x){\sin }^{2}x$

$f(x)=\tan (-x),g(x)=\frac{\tan x-\tan (2x)}{1-\tan x\tan (2x)}$

$\sin (\mathrm{75°})$

$\sin (\mathrm{195°})$

$\cos (\mathrm{165°})$

$\cos (\mathrm{345°})$

$\tan (\mathrm{-15°})$

$\tan (x+\frac{\pi }{4})=\frac{\tan x+1}{1-\tan x}$

$\frac{\tan (a+b)}{\tan (a-b)}=\frac{\sin a\cos a+\sin b\cos b}{\sin a\cos a-\sin b\cos b}$

$\frac{\cos (a+b)}{\cos a\cos b}=1-\tan a\tan b$

$\cos \left(x+y\right)\cos \left(x-y\right)={\cos }^{2}x-{\sin }^{2}y$

$\frac{\cos (x+h)-\cos x}{h}=\cos x\frac{\cos h-1}{h}-\sin x\frac{\sin h}{h}$

$\tan (u+v)=\frac{\tan u+\tan v}{1-\tan u\tan v}$

$\tan (u-v)=\frac{\tan u-\tan v}{1+\tan u\tan v}$

$\frac{\tan \left(x+y\right)}{1+\tan x\tan x}=\frac{\tan x+\tan y}{1-{\tan }^{2}x{\tan }^{2}y}$

If $\alpha ,\beta ,$ and $\gamma$ are angles in the same triangle, then prove or disprove $\sin \left(\alpha +\beta \right)=\sin \gamma .$

If $\alpha ,\beta ,$ and $y$ are angles in the same triangle, then prove or disprove $\tan \alpha +\tan \beta +\tan \gamma =\tan \alpha \tan \beta \tan \gamma$