Question

cos1°×cos2°×cos3°....cos89°

Answer 1

cos1°=cos(90-89)°=sin89°

cos89°=cos(90-1)=sin1°

we make pair of sine series

(sin1°sin89°)(sin2°sin88).......(sin44°sin46°)(sin45)

= 1/√2(sin1°sin89°)(sin2°sin88°)....(sin44°sin46°)

Answer 2

Step-by-step explanation:

To find the value of cos1°×cos2°×cos3°....cos87°×cos88°×cos89°

As we know that

cos1°×cos2°×cos3°....cos(90°-3°)×cos(90°-2°)×cos(90°-1°)

cos1°×cos2°×cos3°...cos 44°×cos45°×sin44°...sin 3°×sin 2°× sin1°

cos1°sin1°×cos2°sin2°×

cos3°sin3°×...cos44°sin44°×cos 45°

we know that

2sin x cos x= sin 2x

To convert the expression,we have to multiply and divide 2 ,44 times

$\frac{2 \sin(1°) \cos(1°) \times2 \sin(2°) \cos(2°) \times 2 \sin(3°) \cos(3°) \times...2 \sin(44°) \cos(44°) \times \: cos45° }{ {2}^{44} } \\ \\ = \frac{ \sin(2°) \times \sin(4°) \times \sin(6°)... \sin(88°) \times \frac{1}{ \sqrt{2} } }{ {2}^{44} } \\ \\ \cos(1°) \cos(2°) ... \cos(88°) \cos(89°) \\\\= \frac{1}{ \sqrt{2} \times {2}^{44} } \sin(2°) \times \sin(4°) \times \sin(6°)... \sin(88°)$

Hope it helps you