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katovenus [111]
2 years ago
6

Answer the question in the picture

Mathematics
1 answer:
nekit [7.7K]2 years ago
4 0

Recall the angle sum identities:

\sin(x+y)=\sin x\cos y+\cos x\sin y

\cos(x+y)=\cos x\cos y-\sin x\sin y

Now,

\tan(x+y)=\dfrac{\sin(x+y)}{\cos(x+y)}=\dfrac{\sin x\cos y+\cos x\sin y}{\cos x\cos y-\sin x\sin y}

Divide through numerator and denominator by \cos x\cos y to get

\tan(x+y)=\dfrac{\tan x+\tan y}{1-\tan x\tan y}

Next, we use the fact that x,y lie in the first quadrant to determine that

\sin x=\dfrac12\implies\cos x=\sqrt{1-\sin^2x}=\dfrac{\sqrt3}2

\cos y=\dfrac{\sqrt2}2\implies\sin x=\sqrt{1-\cos^2x}=\dfrac1{\sqrt2}

So we then have

\tan x=\dfrac{\sin x}{\cos x}=\dfrac{\frac12}{\frac{\sqrt3}2}=\dfrac1{\sqrt3}

\tan y=\dfrac{\sin y}{\cos y}=\dfrac{\frac1{\sqrt2}}{\frac{\sqrt2}2}=1

Finally,

\tan(x+y)=\dfrac{\frac1{\sqrt3}+1}{1-\frac1{\sqrt3}}=\dfrac{1+\sqrt3}{\sqrt3-1}=2+\sqrt3\approx3.73

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<u>Pre-Alg</u>

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