 Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
`        I am value of expression in the image. Please help in finding its vale i am not what to do with this problem. How should i proceed.`
5 months ago

```							The equation cos(9x)=0 has roots x=(2n+1)π/18 for n=0,1,2….17 … (i) But cos(9x) = cos⁹x*{ 9C0 – 9C2*t² + 9C4*t⁴ − 9C6*t⁶ + 9C8*t⁸ } (t=tanx, see end) So either cosx=0 or 9C0 – 9C2*t² + 9C4*t⁴ − 9C6*t⁶ + 9C8*t⁸ = 0  From (i) n=4,13 solve cos(x)=0 and using tan(180−x)=−tan(x) gives  ±tan(10), ±tan(30), ±tan(50), ±tan(70) as solutions of the octic. Further setting z=t² gives tan²(10), tan²(30), tan²(50), tan²(70) as the solutions of 9C0 – 9C2*z + 9C4*z² − 9C6*z³ + 9C8*z⁴ = 0  By root properties tan²(10)+tan²(30)+tan²(50)+tan²(70) = 9C6/9C8 = 84/9 But tan²(30) = (1/√3)² = ⅓ and so tan²(10)+tan²(50)+tan²(70) = 9  You can also get from the same equation … tan(10)*tan(50)*tan(70) = sqrt( (9C0/9C8) ) / tan(30) = 1/√3  cot²(10)+cot²(50)+cot²(70) = 9C2/9C0 – cot²(30) = 33  ………………………………. cos(9x)+isin(9x) = (cosx+isinx)⁹ = ∑ (r=0,9) 9Cr*cos^(9-r)x*(isinx)^r  = cos⁹x * ∑ (r=0,9) 9Cr*(itanx)^r Equating real parts : cos(9x) = cos⁹x * { 9C0 – 9C2*t² + 9C4*t⁴ − 9C6*t⁶ + 9C8*t⁸ } Where t=tanx
```
5 months ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Trigonometry

View all Questions »  ### Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution  ### Course Features

• 31 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions