Question

cos10 cos50 cos70-sin10 sin50 sin70​

Answer 1

HELLO DEAR,

GIVEN:- cos10cos50cos70 - sin10sin50sin70

now, we know:- 2cosa.cosb = Cos(a + b) + cos(a - b)

solving, cos10cos50cos70

=> 1/2{2cos70 + cos10} cos50

=> 1/2 {cos(70 + 10) + cos(70 - 10)} cos50

=> 1/2 {cos80 + cos60} cos50

=> 1/2 {cos80 * cos50} + 1/2{cos60 * cos50}

=> 1/4 {cos(80 + 50) + cos(80 - 50)} + 1/4*cos50

=> 1/4 {cos130 + cos30} + 1/4cos50

=> 1/4 {cos130 + √3/2} + 1/4cos50

=> 1/4 {cos130 + √3/2} + 1/4cos50

=> 1/4 {cos(180 - 50) + √3/2} + 1/4cos50

=> 1/4(-cos50) + √3/8 + 1/4cos50

=> √3/8

similarly, sin10sin50sin70

=> sin10sin50sin70

=> 1/2(2sin10sin50sin70)

we know:- [ 2sinAsinB = cos (A-B) - cos (A+B)]

So,

1/2(2sin10sin50sin70)

=> 1/2[sin10{cos20 - cos120}]

[ cos 120 = cos (90 + 30) = 1/2]

So, 1/2 {sin10(cos20 +1/2)}

=> 1/2sin10cos20 +1/4sin10

=> 1/4(2sin10cos20) + 1/4sin10

As, [2sinAcosB = sin(A + B) - sin(A - B)]

= 1/4 (sin 30 - sin10) + 1/4sin10

= 1/4 sin30

= 1/8

hence, cos10cos50cos70 - sin10sin50sin70

=> √3/8 - 1/8

=> (√3 - 1)/8. <=answer

Answer 2

Answer:

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