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