Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
Here,given 5A and 9A are supplementary angles;
So, 5A+9A=π
14A=π
Then, divided by 2 we get,
7A =π/2,it is also written as;
4A +3A =π/2
4A =π/2 –3A
By taking cosine on both side ,we get
cos(4A)=cos(π/2 −3A)
This equation is satisfied by A=π/14,3π/14,
5π/14 ….
2cos²(2A)−1=sin(3A)
By using formula 1–cos2A =2sin²A
=>2[1–2sin²A]²−1=[3sinA–4sin³3A]
Let x=sinA
=>2(4x⁴−4x²+1)−1=3x–4x³
=>8x⁴−8x²+2–1=3x–4x³
=>8x⁴+4x³−8x²–3x +1=0
=>8x⁴–8x²+4x³–3x+1=0
=>8x²(x²–1)+4x(x²–1) +x+1=0
=>(x²–1)(8x²+4x)+(x+1) =0
=>(x+1)(x–1)(8x²+4x)+(x+1)=0
(x+1){(x–1)(8x²+4x) +1}=0
As you can see at x=−1,the above equation are satisfy and (x+1) is a one of its factor other 3 values of x possible.
So,we consider here only x =–1=sinA
Then,by putting x=sinA, we get the value of
4cos2A–sec4A =4(1–2sin²A) - 1/(cos4A)
Because secA = 1/cosA
=4(1–2sin²A) - 1/(2cos²2A –1)
=4(1–2sin²A) - 1/{2(1–2sin²A)²–1}
=4{1–2(-1)²} - 1/{2(1–2×-1²)²–1}
=4(1–2) - 1/{2(1–2)²–1}
=–4 - 1/(2–1)
=–4 –1
–5,
Hence, the value 4cos2A –sec4A = –5,Ans.
You try to find other values of 4cos2A–sec4A
I hope it helps you…!!!
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !