sin a sin b sin c formula

Thecorrect option is B0 °Consider sin 2 A = 2 sin A.. iWe know that sin 2 A = 2 sin A cos A. i iSubstitute i i in i, we getsin 2 A = 2 sin A ⇒ 2 sin A cos A = 2 sin A ⇒ cos A = 1 b u t cos 0 ° = 1 ⇒ A = 0 °Hence, sin 2 A = 2 sin A is true when A = 0 °. Suggest Corrections. 94. Explanation LH S = sina + sinb + sinc − sin(a +b +c) = sina + sinb + sinc −sin((a + b + c) = 2sin( a + b 2)cos( a − b 2) −2cos( a + b + 2c 2)sin( a + b 2)] = 2sin( a + b 2)[cos( a − b 2) −cos( a + b +2c 2)] = 2sin( a + b 2) × 2sin( a +b + 2c 4 − a −b 4)sin( a +b +2c 4 + a − b 4) Herewe want +1 and as such we write cos2A=1−2sin 2A and combine the other two terms. L.H.S. =1−2sin 2A+2sin(B+C)sin(C−B) (Note) =1−2sin 2A−2sinAsin(B−C) [∵sin(−θ)=−sinθ] =1−2sinA[sinA+sin(B−C)] =1−2sinA[sin(B+C)+sin(B−C)] =1−2sinA(2sinBcosC) =1−4sinAsinBcosC. Solve any question of Trigonometric Functions with:-. 解先把a+b看成一个整体，，那么就有sin（a+b+c）=sin（a+b）cosc+cos（a+b）sinc. 再展开：=（sin a* cosb+cos a*sinb）cos c+（cos a *cos b-sin a*sin b）sinc. =sin a*cos b* cos c+cos a *sin b *cos c +cos a*cos b* sin c- sin a* sinb *sin c. 过程可能有错，但是方法和思路就是这样，你可以换着把b和c contoh surat izin study tour untuk orang tua.

sin a sin b sin c formula