Question

Solve questions 15,16,17, and 18. Will give Brainliest!! Please!!! Show minimal work!

Answer 1

15. The terminal side of $\frac{3\pi}{2}$ intercepted the unit circle at:

(0,-1).

This implies that:

$(\cos \frac{3\pi}{2}, \sin \frac{3\pi}{2}=(0,-1)$

This implies that:

$\cos \frac{3\pi}{2}=0,\:and\: \sin \frac{3\pi}{2}=-1$

16) We have that: $\cot \theta =-\frac{6}{7}$.

We illustrate this on the right triangle and apply the Pythagoras Theorem as follows:

$h^2=6^2+7^2$

$h^2=36+49$

$h^2=85$

$h=\sqrt{85}$

Using the mnemonics SOH-CAH-TOAH, we have:

$\sin \theta=-\frac{Opp}{Hyp}=-\frac{7}{\sqrt{85} }=-\frac{7\sqrt{85} }{85}$

$\cos \theta=\frac{Adj}{Hyp}=\frac{6}{\sqrt{85} }=\frac{6\sqrt{85} }{85}$

$\tan \theta=-\frac{Opp}{Adj}=-\frac{7}{6}}$

$\csc \theta=-\frac{Hyp}{Opp}=-\frac{\sqrt{85} }{7}$

$\sec \theta=\frac{Hyp}{Adj}=\frac{\sqrt{85} }{6}$

17. We want to verify that: $sin^4x-sin^2x=cos^4x-cos^2x$

Verifying from the LHS

$\sin^4x-sin^2x=(sin^2x)^2-sin^2x$

Recall that from the Pythagorean identity:$\sin^2x=1-\cos^2x$

$\sin^4x-sin^2x=(1-cos^2x)^2-(1-\cos^2x)$

$\sin^4x-sin^2x=1-2cos^2x+\cos^4x-1+\cos^2x$

$\sin^4x-sin^2x=\cos^4x-\cos^2x$

18.  We have:

$\tan^2x\sec^2x+2\sec^2x-\tan^2x=2$

$\tan^2x\sec^2x+2\sec^2x-\tan^2x-2=0$

$\sec^2x(\tan^2x+2)-1(\tan^2x+2)=0$

$(\sec^2x-1)(\tan^2x+2)=0$

When $(\sec^2x-1)=0$, we have

$x=0,\pi$

When $\tan^2x+2=0$, $\tan^2x=-2$

We can see this is not defined for all real values of x.