Find the value of (sinθ+cosθ) ²+ (cosθ-sinθ) ².
Answers
Step-by-step explanation:
sin²+cos²+2sin.cos+sin²+cos²-2sin cos
=2sin²+2cos²
=2(sin²+cos²)
=2(1)
=2
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Answer :-
2
Given to find the value of :-
(sinθ+cosθ) ²+ (cosθ-sinθ) ²
SOLUTION:-
Expanding the equation by the algebraic identity (a+b)² and (a-b)²
- (a+b)² = a² + 2ab + b²
- (a-b)² = a² -2ab + b²
= (sinθ)² + (cosθ)² + 2(sinθ)(cosθ) + (cosθ)² + (sinθ)² -2 (sinθ)(cosθ)
= sin²θ+cos²θ + 2sinθcosθ + cos²θ + sin²θ - 2sinθ cosθ
Keeping like terms together
= sin²θ + cos²θ + cos²θ + sin²θ + 2sinθ cosθ - 2sinθ cosθ
= 2 sin²θ + 2 cos²θ
Take common "2 "
= 2 [ sin²θ + cos²θ]
From Trigonometric identities sin²θ + cos²θ = 1
= 2(1)
= 2
So, the value of (sinθ+cosθ) ²+ (cosθ-sinθ) ² is 2
Used Algebraic identities:-
(a+b)² = a² + 2ab + b²
(a-b)² = a² -2ab + b²
Used Trigonmetric Identity :-
sin²θ + cos²θ = 1
Know more Trigonmetric identities , relations , ratios:-
Trigonometric Identities:-
sin²θ + cos²θ = 1
sec²θ - tan²θ = 1
csc²θ - cot²θ = 1
Trigonometric relations:-
sinθ = 1/cscθ
cosθ = 1 /secθ
tanθ = 1/cotθ
tanθ = sinθ/cosθ
cotθ = cosθ/sinθ
Trigonometric ratios:-
sinθ = opp/hyp
cosθ = adj/hyp
tanθ = opp/adj
cotθ = adj/opp
cscθ = hyp/opp
secθ = hyp/adj
Know more Algebraic identities:-
( a + b )² + ( a - b)² = 2a² + 2b²
( a + b )² - ( a - b)² = 4ab
(a+b)(a-b) = a² -b²
( a + b + c )² = a² + b² + c² + 2ab + 2bc + 2ca
a² + b² = ( a + b)² - 2ab
(a + b )³ = a³ + b³ + 3ab ( a + b)
( a - b)³ = a³ - b³ - 3ab ( a - b)
If a + b + c = 0 then a³ + b³ + c³ = 3abc