Question

If , sin theta + root over 3 cos theta then the maximum value of y is

Answer 1

Step-by-step explanation:

y =2/(sinθ + √3cosθ)

for solving this question, you should understand one thing

-\bf{\sqrt{a^2+b^2}} ≤ asinx + bcosx ≤\bf{\sqrt{a^2+b^2}}

So, -√(1 + √3²) ≤ (sinθ + √3cosθ) ≤ √(1 + √3²)

-2 ≤ (sinθ + √3cosθ) ≤ 2

So, minimum value of (sinθ + √3cosθ) = -2

Maximum value of (sinθ + √3cosθ) = 2

For getting minimum value of y , we have to use maximum value of (sinθ + √3cosθ) .

So, minimum value of y = 2/-2 = 1

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Answer 2

Answer:

The maximum value of y = 2.

Step-by-step explanation:

Given  y = sinθ + √3cosθ

Multiplying and dividing by 2 on RHS.

⇒y = 2 [1/2 sinθ + √3/2 cosθ ]

⇒y = 2[cos60° sinθ + sin60°cosθ ]

It is in the form of cosA sinB + sinA cosB = sin(A + B)

so, cos60° sinθ + sin60° cosθ = sin(θ + 60°)

⇒y = 2sin(θ + 60°)

we know the maximum value of the sine function is 1.

then maximum value of sin(θ + 60°) = 1

⇒ y = 2×1 = 2

Therefore the maximum value of y = 2.