Explaining the concept by finding the range of 5 sinx + 12 cosx – 13

Comparing (5 sinx + 12 cosx) with (asinx + bcosx)
a = 5
b = 12
√(a2 + b2) = √(52 + 122) = √(25 + 144) = √169 = 13

asinx + bcosx = √(a2 + b2)(asinx + bcosx) / √(a2 + b2)

Putting values of a & b ==>
5 sinx + 12 cosx = 13 [(5/13)sinx + (12/13)cosx]   …… eq (1)

Since [(5/13)2 + (12/13)2 = 1], comparing with [cos2 a + sin2 a = 1]
So,
sina = 12/13
cosa = 5/13

Puting values in eq (1) –
5 sinx + 12 cosx = 13 [cosa sinx + sina cosx] = 13 sin (x+a)    ….. (2)

Range of sin function is always -1 to 1, so
-1 < sin (x+a) < 1

Multiplying the inequality by 13
-13 < 13 sin (x+a) < 13

By eq (2)-
-13 < (5 sinx + 12 cosx) < 13

Subtracting 13 from the inequality
– 13 – 13 < (5 sinx + 12 cosx – 13) < 13 – 13

-26 < (5 sinx + 12 cosx – 13) < 0

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