Compound Angle Formula

May 2010
8
0
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)
 
Feb 2010
1,036
386
Dirty South
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)
\(\displaystyle cos(A-B) = (cosA \times cosB) + (sinA \times sinB)\)

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

Now try solving your question
 
  • Like
Reactions: coolhacker

Soroban

MHF Hall of Honor
May 2006
12,028
6,341
Lexington, MA (USA)
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:

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

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

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

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

. . \(\displaystyle =\qquad -\cos y \qquad + \qquad 0 \qquad + \qquad 0 \qquad + \qquad \cos y \)

. . \(\displaystyle =\) . . . . . . . . . . . . . . . . . .\(\displaystyle 0\)