Question

5cos x +3 sin(π/6-x)+4 find minimum and maximum value​

Answer 1

Given:

5cos x +3 sin(π/6-x)+4

To find:

Find the minimum and maximum value​.

Solution:

1) You all know that the maximum and minimum value of Sinx and Cosx that is +1 and -1 respectively.

2) So to find the maximum value of the expression put the maximum value in the equation.

• 5cos x +3 sin(π/6-x)+4

• 5.1 + 3.1 +4

• 5+3+4

• 12

3) For minimum value put the minimum value in the equation.

• 5cos x +3 sin(π/6-x)+4

• 5.(-1) + 3.(-1) + 4

• -5 -3 + 4

• -4

The minimum and maximum value is 12 ans -4 respectively.

Answer 2

Answer:

Maximum = 11

Minimum = -3

Step-by-step explanation:

Given: 5cosx + 3sin(π/6 - x) + 4

For finding minimum and maximum value, we're gonna use asinx ± bcosx = ±√a²+b² (+ve for max, -ve for min) , but the equation is not in that form so we'll expand the sine part and then make it so in asinx ± bcosx form like:

Using: $sin(A - B) = sinA•cosB - cosA•sinB$

• 5cosx + 3(sin(π/6)cosx - cos(π/6)sinx) + 4

• cosx(5 + 3sinπ/6) - sinx(3cosπ/6) + 4

Now putting values for standard angles, we get:

• cosx(5 + 3/2) - sinx(3√3/2) + 4

Now we'll just use the formula of asinx ± bcosx

For Maximum:

• 4 + √{(5 + 3/2)² + ((3√3)/2)²}
• 4 + √{169/4 + 27/4}
• 4 + √49 = 11

Similarly, for Minimum:

• 4 - √49 = -3

P.S.- My first answer so pardon me for writing it so poorly :')