Question #244069
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
1
Expert's answer
2021-09-29T17:22:18-0400

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


(sinx+cosx)4=((sinx+cosx)2)2=(sin2x+cos2x+2sinxcosx)2Then we use the relationssin2x+cos2x=1and 2sinxcosx=sin2xto find(sinx+cosx)4=(1+sin2x)2(sinx+cosx)4=1+2sin2x+sin22x(\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 (sinx+cosx)4=1+2sin2x+sin22x(\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 (sinx+cosx)4=0(\sin x+ \cos x)^4=0


(sinx+cosx)4=0(1+sin2x)2=0(1+2sinxcosx)2=01+2sinxcosx=0    sinxcosx=12=12×12(\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<18090<x<180 and that also has to be equal in magnitude, which implies that x has to be 135° since


sin135°×cos135°=12×12=12\sin{135°}\times \cos{135°}=\cfrac{1}{\sqrt{2}}\times-\cfrac{1}{\sqrt{2}}=-\cfrac{1}{2}


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