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Asked by Omer

Express 4sin theta - cos( pi/2 - theta) as a single trigonometric function. Their answer is converts the whole -cos section including the brackets in to -sin theta and I don’t understand why?

Consider the addition formulae: Cos(a+b) = cos(a)cos(b) - sin(a)sin(b) Sin(a+b)=sin(a)cos(b) + sin(b)cos(a) (I'm going to do the following in degrees, so just remember 90 degrees is π/2 rads, and I'll use x instead of theta, again for ease) Now let a=90 and b=x. Then we have  cos(90-x) =  cos(90)cos(-x) - sin(90)sin(-x) Cos(90) = 0 and sin(90) = 1 And cos(-x) = cos(x) [look at the graph of cos(x), it's symmetrical about the y-axis] and sin(-x) = -sin(x) [again, consider the graphs] Therefore we have Cos(90-x) = 0 * cos(x) - 1 * -sin(x) = sin(x) Hence your question sinplifies to 4sin(theta) - cos(π/2 - theta) = 4sin(theta) - sin(theta) = 3sin(theta) Hope this helps.

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Douglas McGuire
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