arccos(sin(-4)) looking for step by step solution using unit circle

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arccos(sin(-4)) how to find this answer

Aug 20th, 2015

Hi there! Thank you for the opportunity to help you with your question!

Another way of writing sin(-4) = cos(-4+3*pi/2) 

We can check this by doing the angle addition formula for cosine:

cos(a+b) = cos(a)cos(b) - sin(a)sin(b)

In this case:

cos(-4+3pi/2) = cos(-4)cos(3pi/2) - sin(-4)sin(3pi/2).

From the unit circle, we know that: cos(3pi/2) = 0 and sin(3pi/2) = -1, we recover that 

cos(-4+3pi/2) = sin(-4) 

Therefore, when we take the arccos of both sides we get:

arccos( cos(-4+3pi/2)) = arccos(sin(-4))

And on the left hand side, the arccos and the cos cancel giving:

-4 + 3pi/2 = arccos(sin(-4))

*Note: the range of the arccos function is from -1 to 1, therefore we need to make sure that the output we get is in that range. In this case, we see that -4+3pi/2 = 0.712..., which is between -1 and 1. So final answer is -4+3pi/2


Please let me know if you need any clarification. Always glad to help!
Aug 20th, 2015

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