3sinx + 4cosx= 5 Find sinx?
3sin(x) + 4cos(x) = 5
(3/5)sin x + (4/5) cos x = 1
put (3/5) = cos y and (4/5) = sin y, because sin^2 y +cos^2 y = 16/25+9/25=25/25=1
sin x(cos y) +cos x (sin y) = 1
sin (x+y) = 1 = sin 90
x + y = 90 or pi/2
x = (90- y) or x = (pi/2 - y)
therefore sin x = cos y = 3/5
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3sinx + 4cosx= 5 Find sinx?
Given Equation: 3sin(x) + 4cos(x) = 5
To find the value of sin(x), we can rearrange the equation using trigonometric identities and solve for sin(x).
Rewriting the Equation:
We know that sin^2(x) + cos^2(x) = 1 (Identity 1)
Dividing the entire equation by 5, we get:
(3/5)sin(x) + (4/5)cos(x) = 1
Now, let's introduce a new variable, let's say θ, such that sin(θ) = (3/5) and cos(θ) = (4/5).
Using the Pythagorean identity sin^2(θ) + cos^2(θ) = 1, we can verify that θ is a valid angle.
Using Trigonometric Identities:
We can rewrite the equation using the values of sin(θ) and cos(θ):
(3/5)sin(x) + (4/5)cos(x) = 1
(3/5)sin(x) + (4/5)cos(x) = sin^2(θ) + cos^2(θ)
(3/5)sin(x) + (4/5)cos(x) = sin^2(θ) + cos^2(θ) + 2sin(θ)cos(θ) - 2sin(θ)cos(θ)
(3/5)sin(x) + (4/5)cos(x) = sin^2(θ) + 2sin(θ)cos(θ) + cos^2(θ) - 2sin(θ)cos(θ)
(3/5)sin(x) + (4/5)cos(x) = sin^2(θ) + cos^2(θ) + 2sin(θ)cos(θ) - 2sin(θ)cos(θ)
(3/5)sin(x) + (4/5)cos(x) = sin^2(θ) + 2sin(θ)cos(θ) + cos^2(θ) - 2sin(θ)cos(θ)
(3/5)sin(x) + (4/5)cos(x) = (sin(θ) + cos(θ))^2 - 2sin(θ)cos(θ)
(3/5)sin(x) + (4/5)cos(x) = (sin(θ) + cos(θ))^2 - 2sin(θ)cos(θ)
Using the Given Values:
Now, substituting the values of sin(θ) = (3/5) and cos(θ) = (4/5):
(3/5)sin(x) + (4/5)cos(x) = (3/5 + 4/5)^2 - 2(3/5)(4/5)
(3/5)sin(x) + (4/5)cos(x) = (7/5)^2 - 2(3/5)(4/5)
(3/5)sin(x) + (4/5)cos(x) = (49/25) - (24/25)
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3sinx + 4cosx= 5 Find sinx?
3 sin x + 4 cos x= 5 or,(3 sin x + 4 cos x)/5=1 or,sin x =(4 cos x /5)×4/3or, sin x= 3/5
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