Answer to Question #244069 in Trigonometry for Adawiyah

Question

Prove that (sinx+cosx)^4 = sin^2 2x+ 2sin2x+1 where 90<x<180 . Hence, find the value of x if (sinx + cosx)^4=0

Answer 1

We have to use the trigonometric equivalences for the sum of angles and the pitagoric relation to find:

$(\sin x+ \cos x)^4=\Big((\sin x+ \cos x)^2 \Big)^2 \\=(\sin^2 x+ \cos^2 x+2 \sin x \cos x )^2 \\ \text{Then we use the relations} \\ \sin^2 x+\cos^2 x=1 \\ \text{and } 2 \sin x \cos x =\sin{2x} \\ \text{to find} \\ (\sin x+ \cos x)^4= (1+\sin {2x} )^2 \\ (\sin x+ \cos x)^4= 1+2\sin{2x}+\sin ^2{2x}$

Then, after we prove that $(\sin x+ \cos x)^4= 1+2\sin{2x}+\sin ^2{2x}$, we use one of the relation to find the value of x that safisfies $(\sin x+ \cos x)^4=0$

$(\sin x+ \cos x)^4=0 \\ (1+\sin {2x})^2=0 \\ \sqrt {(1+2\sin {x}\cos x)^2}=0 \\ 1+2\sin {x}\cos x=0 \\\implies \sin {x}\cos x=-\frac{1}{2}=-\frac{1}{\sqrt{2}}\times\frac{1}{\sqrt{2}}$

Now, since we need a value for x that implies a positive value for the sine and a negative value for the cosine under $90<x<180$ and that also has to be equal in magnitude, which implies that x has to be 135° since

$\sin{135°}\times \cos{135°}=\cfrac{1}{\sqrt{2}}\times-\cfrac{1}{\sqrt{2}}=-\cfrac{1}{2}$