Hello, jas397!
Don't you know
ANY of the basic trig facts?
\(\displaystyle [1]\;\sin x\,=\,\frac{1}{\csc x},\;[2]\;\csc x = \frac{1}{\sin x}\)
\(\displaystyle [3]\;\cos x\,=\,\frac{1}{\sec x},\;[4]\;\sec x\,=\,\frac{1}{\cos x}\)
\(\displaystyle [5]\;\tan x\,=\,\frac{1}{\cot x},\;[6]\;\cot x\,=\,\frac{1}{\tan x}\)
\(\displaystyle [7]\;\tan x\,=\,\frac{\sin x}{\cos x},\;[8]\;\cot x\,=\,\frac{\cos x}{\sin x}\)
\(\displaystyle [9]\;\sin^2x\,+\,\cos^2x\:=\:1,\;[10]\;\tan^2x\,+\,1\:=\:\sec^2x,\;[11]\;\cot^2x\,+\,1\:=\:\csc^2^2x\)
Prove each identity.
\(\displaystyle a)\;\sin(x)\cdot\sec(x)\,=\,\tan(x)\)
We have: \(\displaystyle \,\sin(x)\cdot\sec(x)\)
From [4], this becomes: \(\displaystyle \,\sin(x)\cdot\frac{1}{\cos x}\:= \:\frac{\sin(x)}{\cos(x)}\)
From [7], we have: \(\displaystyle \.\frac{\sin(x)}{\cos(x)}\:=\:\tan(x)\)
\(\displaystyle b)\;\cos^2(x)\,-\,sin^2(x)\:=\:1\,-\,2\cdot\sin^2(x)\)
The left side is: \(\displaystyle \,\cos^2(x)\,-\,\sin^2(x)\)
\(\displaystyle \;\;\;\)From [9], we have: \(\displaystyle \sin^2x\,+\,\cos^2x\:=\:1\;\;\Rightarrow\;\;cos^2x\:=\:1\,-\,\sin^2x\)
Substitute: \(\displaystyle \,(1\,-\,\sin^2x)\,-\,\sin^2x\:=\:1\,-\,2\sin^2x\)
See how it's done?
