Find cos248o - sin212o , if sin18o = (√5 - 1)/4
Let's solve the problem step by step:
1. Express cos 248° and sin 212° in terms of known angles:
* cos 248° = cos (180° + 68°) = -cos 68°
* sin 212° = sin (180° + 32°) = -sin 32°
So the expression becomes:
cos 248° - sin 212° = -cos 68° - (-sin 32°) = -cos 68° + sin 32°
2. Express cos 68° and sin 32° in terms of sin 18°:
* cos 68° = cos (90° - 22°) = sin 22°
* sin 32° = sin (18° + 14°)
We will use the following trigonometric identities:
* sin (a + b) = sin a cos b + cos a sin b
* sin (2a) = 2 sin a cos a
* cos (2a) = 1 - 2 sin^2 a
* cos (a) = sqrt(1-sin^2(a))
3. Express sin22 in terms of sin18
Sin(22) = sin(18+4)
Sin(22) is difficult to express directly in terms of sin18. To solve this problem a different approach is needed.
4. Alternative Approach:
We can rewrite the expression as:
cos 248° - sin 212° = -cos 68° + sin 32°
We can rewrite the expression as:
-cos(90-22) + sin(18+14) = -sin(22) + sin(18+14)
We can also write:
-cos(270-22) - sin(180+32) = -sin(22)+sin(32)
Lets use sin(32) = sin(2*16) and sin(22)= sin(2*11)
cos(248) - sin(212) = -cos(68) + sin(32)
We can express cos(68) as cos(90-22) = sin(22)
Then we have, -sin(22) + sin(32)
sin(32) - sin(22)
Use sin(A) - sin(B) = 2 cos((A+B)/2)sin((A-B)/2)
sin(32) - sin(22) = 2 cos(27) sin(5)
5. Express cos 27 and sin 5 in terms of sin18
This is very difficult to do. We are going to use a calculator to find the numeric values.
cos 248° ≈ -0.3746
sin 212° ≈ -0.5299
cos 248° - sin 212° ≈ -0.3746 - (-0.5299) ≈ 0.1553
Using a calculator:
sin(18) = (sqrt(5)-1)/4 ≈ 0.3090.
2cos(27)sin(5) ≈ 0.1553
Therefore, the value is approximately 0.1553.
Final Answer: The final answer is approximately 0.1553.
The elementary properties of inverse trigonometric functions will help to solve problems. Here are a few important properties related to inverse trigonometric functions:
Tan−1x + Tan−1y = π + tan−1 (x+y/ 1-xy), if xy > 1
Tan−1x + Tan−1y = tan−1 (x+y/ 1-xy), if xy < 1
Tan−1x + Tan−1y = tan−1 (x+y/ 1-xy), if xy < 1
Tan−1x + Tan−1y = -π + tan−1 (x+y/ 1-xy), if xy > 1
= x, if x∈[−π/2, π/2]
= π−x, if x∈[π/2, 3π/2]
=−2π+x, if x∈[3π/2, 5π/2] And so on.
= −x, ∈[−π,0]
= x, ∈[0,π]
= 2π−x, ∈[π,2π]
=−2π+x, ∈[2π,3π]
= x, (−π/2, π/2)
= x−π, (π/2, 3π/2)
= x−2π, (3π/2, 5π/2)