15. The terminal side of [tex]\frac{3\pi}{2}[/tex] intercepted the unit circle at:
(0,-1).
This implies that:
[tex](\cos \frac{3\pi}{2}, \sin \frac{3\pi}{2}=(0,-1)[/tex]
This implies that:
[tex]\cos \frac{3\pi}{2}=0,\:and\: \sin \frac{3\pi}{2}=-1[/tex]
16) We have that: [tex]\cot \theta =-\frac{6}{7}[/tex].
We illustrate this on the right triangle and apply the Pythagoras Theorem as follows:
[tex]h^2=6^2+7^2[/tex]
[tex]h^2=36+49[/tex]
[tex]h^2=85[/tex]
[tex]h=\sqrt{85}[/tex]
Using the mnemonics SOH-CAH-TOAH, we have:
[tex]\sin \theta=-\frac{Opp}{Hyp}=-\frac{7}{\sqrt{85} }=-\frac{7\sqrt{85} }{85}[/tex]
[tex]\cos \theta=\frac{Adj}{Hyp}=\frac{6}{\sqrt{85} }=\frac{6\sqrt{85} }{85}[/tex]
[tex]\tan \theta=-\frac{Opp}{Adj}=-\frac{7}{6}}[/tex]
[tex]\csc \theta=-\frac{Hyp}{Opp}=-\frac{\sqrt{85} }{7}[/tex]
[tex]\sec \theta=\frac{Hyp}{Adj}=\frac{\sqrt{85} }{6}[/tex]
17. We want to verify that: [tex]sin^4x-sin^2x=cos^4x-cos^2x[/tex]
Verifying from the LHS
[tex]\sin^4x-sin^2x=(sin^2x)^2-sin^2x[/tex]
Recall that from the Pythagorean identity:[tex]\sin^2x=1-\cos^2x[/tex]
[tex]\sin^4x-sin^2x=(1-cos^2x)^2-(1-\cos^2x)[/tex]
[tex]\sin^4x-sin^2x=1-2cos^2x+\cos^4x-1+\cos^2x[/tex]
[tex]\sin^4x-sin^2x=\cos^4x-\cos^2x[/tex]
18. We have:
[tex]\tan^2x\sec^2x+2\sec^2x-\tan^2x=2[/tex]
[tex]\tan^2x\sec^2x+2\sec^2x-\tan^2x-2=0[/tex]
[tex]\sec^2x(\tan^2x+2)-1(\tan^2x+2)=0[/tex]
[tex](\sec^2x-1)(\tan^2x+2)=0[/tex]
When [tex](\sec^2x-1)=0[/tex], we have
[tex]x=0,\pi[/tex]
When [tex]\tan^2x+2=0[/tex], [tex]\tan^2x=-2[/tex]
We can see this is not defined for all real values of x.
