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- StudyBlue
- Minnesota
- University of Minnesota - Twin Cities
- Mathematics
- Mathematics 1151
- Lyubeznik
- Lecture Number 12: Section 7.4

Linli F.

Lecture Number 12 February 16, 2009 Section 7.4 Cosine: Cos(A + B) = cosAcosB – sinAsinB Cos(A – B) = cosAcosB – sinAsinB The second formula implies the first: Cos(A + B) = cos(A – ( - B) ) = cosAcos(-B) + sinAsin(-B) cos(-B) = cos(B) and sin(-B) = -sin(B) = cosAcosB - sinAsinB Sine: Sin(A + B) = sinAcosB + cosAsinB Sin(A – B) = sinAcosB – cosAsinB Proof: Sin(A + B) = cos (π/2 – (A + B)) = (π/2 – A) – B = cos( ( π/2 – A ) – B ) Tangent: Tan(A + B) = Tan(A – B) = Example) Use one of the equations above to solve sin(π/12) The first step is to change π/12 into one of the equations we’ve previously learned such as π, π/2, π/4, π/3, π/6, or 2 π. Well, looking at π/12, it is the same as (π/3 – π/4). Therefore, the equation would be the same as solving sin(π/3 – π/4). After we do this, we need to transform it into one of the equations above. Well, π/3 would be A and π/4 would be B and it would fit into the equation Sin(A – B) = sinAcosB – cosAsinB. Therefore, it would transform to sin(π/3)cos(π/4) – cos(π/3)sin(π/4). Now, we already know the sine and cosine to π/3 and π/4 so we would change the equation to the actual values and it would look like: . Now, we would just solve by multiplying and subtracting and the answer would end up being:

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