Question

a cosx - b sinx = c, then show that a sinx = b cosx = + or - ( a^2 + b^2 - c^2) ^1/2.

Answer 1

sorry a change
its show that \[a sinx + bcosx= \sqrt{a^2+b^2-c^2}\]

Answer 2

the RHS should be + or -

Answer 3

\[ a \cos x - b \sin x = c \] Squaring both sides,
\( ( a \cos x - b \sin x )^2 = c^2 \) \( \implies a^2(1-\sin^2 x) + b^2(1-\cos^2 x) - 2ab \sin x \cos x = c^2\) \( \implies a^2 - a^2 \sin^2 x +b ^2 - b^2 \cos^2 x - 2ab \sin x \cos x = c^2 \)

Answer 4

\( \implies a^2+b^2-c^2 = ( a \sin x + b \cos x )^2 \)

\( \implies a \sin x + b \cos x = \sqrt{a^2+b^2-c^2} \)

Answer 5

square on both sides of acosx=bsinx+c
\[a ^{2}\cos ^{2}x=b ^{2}\sin ^{2}x+2bcsinx+c ^{2}\]
use \[\cos ^{2}x=1-\sin ^{2}x\]
=> \[(a ^{2}+b ^{2})\sin ^{2}x+2bcsinx+c ^{2}-a ^{2}=0\]
solve the quadratic equation to get sinx and similarly cosx
substitute in the equation asinx +bcosx to get R.H.S