What is cos(arctan(−2)+arccos(513))?

2 Answers
Lovecraft
Sep 24, 2015

cos(arctan(−2)+arccos(513))=29√565

Explanation:

First we expand that:
cos(arctan(−2)+arccos(513))=
cos(arctan(−2))cos(arccos(513))−sin(arctan(−2))sin(arccos(513))

We know that cos(arccos(θ))=θ so we can rewrite that to
5cos(arctan(−2))13−sin(arctan(−2))sin(arccos(513))

We know that sin2(θ)=1−cos2(θ), so
sin2(arccos(513))=1−(513)2=1−25169=144169

Taking the root
sin(arccos(513))=1213
It's positive because the sine is always positive in the arccosine range. It can let us rewrite the first expansion to:
5cos(arctan(−2))13−12sin(arctan(−2))13

Now, from sin2(θ)+cos2(θ)=1, if we divide both sides by cos2(θ) we get tan2(θ)+1=1cos2(θ). So:
tan2(arctan(−2))+1=1cos2(θ)→cos2(θ)=1(−2)2+1
cos2(θ)=15→cos(θ)=1√5=√55

It's positive because on the range of the arctangent, the cosine is always postive. So we can rewrite it to:
√513−12sin(arctan(−2))13

Since the tangent is negative, and the cosine positive, it means the sine is negative. Using sin2(θ)+cos2(θ)=1 we have that
sin2(arctan(−2))=1−15=45
sin(arctan(−2))=−2√5=−2√55

Which means we can rewrite it all to:
√513−12(−2√55)13

From there on it's algebra:
5√565+24√565=
29√565

Nghi N.
Sep 25, 2015

Find: cos(arctan(−2)+arccos(513))

Ans: 1

Explanation:

Use calculator.
tanx=(−2)--> arc x=−63.43 deg
cosy=513=0.38 --> arc y=67.38 deg
cos (-63.43 + 67.58) = cos 3.95 = 0.997 = 1

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