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Grade: 11
        
If 5A and 9A are supplementary angles then find the exact value of (4 cos2A - sec4A)
one month ago

Answers : (2)

Saurabh Koranglekar
askIITians Faculty
8380 Points
							Dear student

5A +9A = 180
A= 180/14
Or pi/14
Exact value of given equation is 2

Regards
one month ago
Vikas TU
11761 Points
							

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…!!!

one month ago
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