Compound Angle Formula

**coolhacker** · May 11th 2010, 07:07 AM

Hi everyone!

plz help me solve this Compound Angle Formula question:

1. Using Compound Angle Formula, prove that:

cos(y - 180°) + sin (y + 90°) = 0

plz thxx

**harish21** · May 11th 2010, 07:32 AM

Originally Posted by coolhacker

Hi everyone!

plz help me solve this Compound Angle Formula question:

1. Using Compound Angle Formula, prove that:

cos(y - 180°) + sin (y + 90°) = 0

plz thxx

$cos(A-B) = (cosA \times cosB) + (sinA \times sinB)$

$sin(A+B) = (sinA \times cosB) + (cosA \times sinB)$

Now try solving your question

**Soroban** · May 11th 2010, 07:37 AM

Hello, coolhacker!

We are expected to know these formulas:

. . $\cos(A-b) \;=\;\cos A\cos B + \sin A\sin B$

. . $\sin(A + B) \;=\;\sin A\cos B + \cos A\sin B$

1. Using Compound Angle Formulas, prove that:

. . . . . $\cos(y - 180^o) + \sin (y + 90^o) \;=\;0$

We have: . $\cos(y-180^o) \qquad\quad+ \quad\qquad\sin(y+90^o)$

. . $=\;\;\overbrace{\cos y \cos180^o + \sin y \sin180^o} + \overbrace{\sin y\cos90^o + \cos y\sin90^o}$

. . $=\;\;\;\cos y\,(-1) \quad+\quad \sin y\,(0) \quad+\quad \sin y\,(0) \quad+ \quad \cos y\,(1)$

. . $=\qquad -\cos y \qquad + \qquad 0 \qquad + \qquad 0 \qquad + \qquad \cos y$

. . $=$ . . . . . . . . . . . . . . . . . . $0$

**coolhacker** · May 11th 2010, 07:56 AM

thx u

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