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Contents
Examples 1 2
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Examples
Now
that we have introduced and proved all the formulae, here are some examples to
reinforce your understanding.
 Example
1:In
the triangle ABC, angle B is obtuse, sin A = 4/5 and sin B = 5/13. Without
using a calculator, find (i) cos C and (ii) tan A/2
Solution: (i)
C = 180o - (A + B)
cos
C = cos [180o - (A + B)]
= -cos (A +
B)
supplementary
angles
= - (cosA cosB - sinA sinB)
addition
formulae
= - [3/5 x (-12/13) - 4/5 x 5/13]
= 56/65
(ii)
tan A = 4/3
4 (1 - tan2 A/2) = 6 tan
A/2
double
angle formulae
2 tan2 A/2 + 3 tan A/2 - 2 = 0
(2 tan A/2 - 1) (tan A/2 + 2) = 0
tan A/2 = 1/2 or tan A/2 = -2
As
A is acute, tan A/2 = 1/2
Example
2:By
expressing 4 sin 2x - 3 cos 2x in the form R sin (2x - k), where R>0 and k
is acute,
(i)
obtain the maximum value of 4 sin 2x - 3 cos 2x
(ii)
solve the equation 4 sin 2x - 3 cos 2x + 2.5 = 0 for all values of x between 0o
and 360o.
Solution:4
sin 2x - 3 cos 2x = (32 + 42)1/2 sin (2x -
tan-1 3/4)
= 5 sin (2x - 36.870o)
(i)
The maximum value of the expression is 5.
(ii)
4 sin 2x - 3 cos 2x + 2.5 = 0
5 sin (2x - 36.870o) = -2.5
sin (2x - 36.870o) = -0.5
2x - 36.870o = -30o, 210o, 330o,
570o
x = 3.4o, 123.4o, 183.4o, 303.4o
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