Topic :- Trigonometry
1) Value of 'cos20° + cos80° - √3.cos50°'.
2) Value of 'cos0° + cosπ/7 + cos2π/7 + cos3π/7 + cos4π/7 + cos5π/7 + cos6π/7'.
3) cos20° + cos40° + cos60° - 4cos10°cos20°cos30°
4) cos20°cos100° + cos100°cos140°+cos140°cos200°
Question from class 9th Trigonometry.
Answers
EXPLANATION.
Question = 1.
⇒ Value of cos20° + cos80° - √3cos50°.
As we know that,
Formula of :
⇒ cos(C) + cos(D) = 2cos[C + D]/2. cos[C - D]/2.
Using this formula in the equation, we get.
⇒ cos(20°) + cos(80°) = 2cos[80° + 20°]/2 . cos[80° - 20°]/2.
⇒ 2cos[100°/2]. cos[60°/2].
⇒ 2cos(50°). cos(30°).
⇒ 2cos(50°).cos(30°) - √3cos50°.
As we know that,
Formula of :
⇒ cos30° = √3/2.
⇒ 2cos(50°) x (√3/2) - √3cos50°.
⇒ √3cos50° - √3cos50° = 0.
⇒ Value of cos20° + cos80° - √3cos50° = 0.
Question = 2.
⇒ Value of cosπ/7 + cos2π/7 + cos3π/7 + cos4π/7 + cos5π/7 + cos6π/7.
As we know that,
⇒ cos(π - θ) = - cosθ.
Using this formula in the equation, we get.
⇒ cosπ/7 + cos2π/7 + cos3π/7 + cos(π - 3π/7) + cos(π - 2π/7) + cos(π - π/7).
⇒ cos(π/7) + cos(2π/7) + cos(3π/7) - cos(3π/7) - cos(2π/7) - cos(π/7).
⇒ cos(0°) = 1.
Question = 3.
⇒ cos20° + cos40° + cos60° - 4cos10°. cos20°. cos30°.
As we know that,
Formula of :
⇒ cos(C) + cos(D) = 2cos[C + D]/2. cos[C - D]/2.
⇒ 2cos(A). cos(B) = cos(A + B) + cos(A - B).
Using this formula in the equation, we get.
⇒ cos20° + cos40° = 2[cos(40° + 20°)/2.cos(40° - 20°)/2].
⇒ 2cos(60°/2). cos(20°/2).
⇒ 2cos(30°).cos(10°).
⇒ cos(10°).cos(20°) = [cos(20° + 10°) + cos(20° - 10°)]/2.
⇒ [cos(30°) + cos(10°)]/2.
We can write equation as,
⇒ 2cos(30°).cos(10°) + cos(60°) - 4[cos(30°) + cos(10°)]/2. cos(30°).
⇒ 2cos(30°).cos(10°) + cos(60°) - 2[cos(30°) + cos(10°)]. cos(30°).
As we know that,
Formula of :
⇒ cos(30°) = √3/2.
⇒ cos(60°) = 1/2.
Using this formula in the equation, we get.
⇒ 2 x (√3/2).cos(10°) + 1/2 - 2[√3/2 + cos(10°)] . (√3/2).
⇒ √3cos(10°) + 1/2 - 2 x (√3/2) [√3/2 + cos(10°)].
⇒ √3cos(10°) + 1/2 - √3[√3/2 + cos(10°)].
⇒ √3cos(10°) + 1/2 - √3 x √3/2 - √3cos(10°).
⇒ 1/2 - 3/2.
⇒ (1 - 3)/2 = - 2/2 = - 1.
⇒ cos20° + cos40° + cos60° - 4cos10°. cos20°. cos30° = - 1.
Question = 4.
⇒ cos(20°).cos(100°) + cos(100°).cos(140°) + cos(140°).cos(200°).
As we know that,
Formula of :
⇒ 2cos(A). cos(B) = cos(A + B) + cos(A - B).
Using this formula in the equation, we get.
⇒ cos(20°).cos(100°) = [cos(100° + 20°) + cos(100° - 20°)]/2.
⇒ cos(20°).cos(100°) = [cos(120°) + cos(80°)]/2.
⇒ cos(100°).cos(140°) = [cos(140° + 100°) + cos(140° - 100°)]/2.
⇒ cos(100°).cos(140°) = [cos(240°) + cos(40°)]/2.
⇒ cos(140°).cos(200°) = [cos(200° + 140°) + cos(200° - 140°)]/2.
⇒ cos(140°).cos(200°) = [cos(340°) + cos(60°)]/2.
We can write equation as,
⇒ [cos(120°) + cos(80°)]/2 + [cos(240°) + cos(40°)]/2 + [cos(340°) + cos(60°)]/2.
⇒ 1/2 [ cos(120°) + cos(80°) + cos(240°) + cos(40°) + cos(340°) + cos(60°)].
As we know that,
Formula of :
⇒ cos(120°) = - 1/2.
⇒ cos(240°) = - 1/2.
⇒ cos(60°) = 1/2.
Using this formula in the equation, we get.
⇒ 1/2 [ -1/2 + cos(80°) + (-1/2) + cos(40°) + cos(340°) + 1/2].
⇒ 1/2 [ cos(80°) + cos(40°) + cos(340°) - 1/2].
As we know that,
Formula of :
⇒ cos(C) + cos(D) = 2cos[C + D]/2. cos[C - D]/2.
Using this formula in the equation, we get.
⇒ 1/2 [ 2cos(80° + 40°)/2. cos(80° - 40°)/2 + cos(340°) - 1/2].
⇒ 1/2 [ 2cos(120°)/2 . cos(40°)/2 + cos(340°) - 1/2].
⇒ 1/2 [ 2cos(60°). cos(20°) + cos(180° x 2 - 20°) - 1/2].
⇒ 1/2 [ 2 x 1/2 . cos(20°) - cos(20°) - 1/2.].
⇒ 1/2 [ cos(20°) - cos(20°) - 1/2].
⇒ 1/2 [ - 1/2].
⇒ - 1/4.
⇒ cos(20°).cos(100°) + cos(100°).cos(140°) + cos(140°).cos(200°) = - 1/4.
Answer:
Consider the given equation
Cos 20° · Cos40° · Cos60°. Cos80° = 1/16
LHS = Cos 20° · Cos40° · Cos60° . Cos80°
We know that Cos60° = 1/2
LHS = Cos 20° · Cos40° · 1/2 . Cos80°
Multiply and divide the equation by 2
LHS = 1/4 (2. Cos 20° · Cos40° · Cos80°)
We know the formula 2 cosa cosb= cos(a+b) + cos(a-b)
LHS = 1/4 [Cos(20+80)+ Cos(20-80)] . Cos40
LHS = 1/4 [Cos(-60)+ Cos(100)] Cos40
LHS = 1/4 [1/2 + cos100] Cos40
LHS = 1/8 Cos40+ 1/4 (Cos40 . Cos100)
Multiply and divide the equation by 2
LHS = 2/2 (1/8 Cos 40) + 1/8(2 Cos40 Cos100)
We know the formula
2cosa cosb= cos(a+b) cos(a-b)
LHS = 1/8 Cos40+ 1/8 [Cos 140 + Cos (-60)]
LHS = 1/8 Cos 40+ 1/8 Cos 140 + 1/16
Since Cos 60= 1/2
LHS = 1/8 (Cos 40 + Cos 140) + 1/16
LHS = 1/8 [2 Cos 90 Cos (-50)] + 1/16
LHS = Cos 90
Cos 90 = 0
LHS = 1/16
LHS = RHS