C coolhacker May 2010 8 0 May 11, 2010 #1 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 (Clapping)(Talking)

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 (Clapping)(Talking)

harish21 Feb 2010 1,036 386 Dirty South May 11, 2010 #2 coolhacker said: 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 (Clapping)(Talking) Click to expand... \(\displaystyle cos(A-B) = (cosA \times cosB) + (sinA \times sinB)\) \(\displaystyle sin(A+B) = (sinA \times cosB) + (cosA \times sinB)\) Now try solving your question Reactions: coolhacker

coolhacker said: 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 (Clapping)(Talking) Click to expand... \(\displaystyle cos(A-B) = (cosA \times cosB) + (sinA \times sinB)\) \(\displaystyle sin(A+B) = (sinA \times cosB) + (cosA \times sinB)\) Now try solving your question

S Soroban MHF Hall of Honor May 2006 12,028 6,341 Lexington, MA (USA) May 11, 2010 #3 Hello, coolhacker! We are expected to know these formulas: . . \(\displaystyle \cos(A-b) \;=\;\cos A\cos B + \sin A\sin B\) . . \(\displaystyle \sin(A + B) \;=\;\sin A\cos B + \cos A\sin B\) 1. Using Compound Angle Formulas, prove that: . . . . . \(\displaystyle \cos(y - 180^o) + \sin (y + 90^o) \;=\;0\) Click to expand... We have: . \(\displaystyle \cos(y-180^o) \qquad\quad+ \quad\qquad\sin(y+90^o)\) . . \(\displaystyle =\;\;\overbrace{\cos y \cos180^o + \sin y \sin180^o} + \overbrace{\sin y\cos90^o + \cos y\sin90^o} \) . . \(\displaystyle =\;\;\;\cos y\,(-1) \quad+\quad \sin y\,(0) \quad+\quad \sin y\,(0) \quad+ \quad \cos y\,(1) \) . . \(\displaystyle =\qquad -\cos y \qquad + \qquad 0 \qquad + \qquad 0 \qquad + \qquad \cos y \) . . \(\displaystyle =\) . . . . . . . . . . . . . . . . . .\(\displaystyle 0\) Reactions: coolhacker and masters

Hello, coolhacker! We are expected to know these formulas: . . \(\displaystyle \cos(A-b) \;=\;\cos A\cos B + \sin A\sin B\) . . \(\displaystyle \sin(A + B) \;=\;\sin A\cos B + \cos A\sin B\) 1. Using Compound Angle Formulas, prove that: . . . . . \(\displaystyle \cos(y - 180^o) + \sin (y + 90^o) \;=\;0\) Click to expand... We have: . \(\displaystyle \cos(y-180^o) \qquad\quad+ \quad\qquad\sin(y+90^o)\) . . \(\displaystyle =\;\;\overbrace{\cos y \cos180^o + \sin y \sin180^o} + \overbrace{\sin y\cos90^o + \cos y\sin90^o} \) . . \(\displaystyle =\;\;\;\cos y\,(-1) \quad+\quad \sin y\,(0) \quad+\quad \sin y\,(0) \quad+ \quad \cos y\,(1) \) . . \(\displaystyle =\qquad -\cos y \qquad + \qquad 0 \qquad + \qquad 0 \qquad + \qquad \cos y \) . . \(\displaystyle =\) . . . . . . . . . . . . . . . . . .\(\displaystyle 0\)