maths extended trigonometry. trigonometrical identities?
Solve for value of x between 0 degrees and 360 degrees
a) 2sinx=cos(x+60)
b) cos(x+45)=cosx
c) sin(x-30)=1/2cosx
d) 3sin(x+10)=4cos(x-10)
Solve for value of x between 0 degrees and 360 degrees
a) 2sinx=cos(x+60)
b) cos(x+45)=cosx
c) sin(x-30)=1/2cosx
d) 3sin(x+10)=4cos(x-10)
Ed I
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a) 2 sin x = cos (x + 60°) = cos x cos 60° - sin x sin 60° = 1/2 cos x - √3/2 sin x
(2 + √3/2) sin x = 1/2 cos x
tan x = 1/(4 + √3)
x ≈ 9.90° or 189.90°
b) cos (x + 45) = cos x
cos x cos 45° - sin x sin 45° = cos x
cos x (√2/2 - 1) = √2/2 sin x
(√2/2 - 1)/(√2/2) = tan x
1 - √2 = tan x
x = 157.5° or 337.5°
c) sin (x - 30°) = 1/2 cos x
sin x cos 30° - cos x sin 30° = 1/2 cos x
√3/2 sin x = cos x
tan x = 2/√3
x ≈ 49.1°, 229.1°
d) 3 sin (x + 10°) = 4 cos (x - 10°)
3 sin x cos 10° + 3 cos x sin 10° = 4 cos x cos 10° + 4 sin x sin 10°
(3 cos 10° - 4 sin 10°) sin x = (4 cos 10° - 3 sin 10°) cos x
tan x = (4 cos 10° - 3 sin 10°)/(3 cos 10° - 4 sin 10°)
tan x = 1.5126...
x ≈ 56.5°, 336.5°
mohanrao d
a) 2sinx=cos(x+60)
=> 2 sin x = (1/2)cos x - ( â3/2) sin x
multiply with 2
=> 4 sin x + â3 sin x = cos x
sin x (4 + â3) = cos x
divide by cos x
tan x (4 + â3) = 1
tan x = 1 /(4 + â3) = (4 -â3) /13
x = 9.9° and 189.9 °
b) cos(x+45) = cosx
=> cos(x + 45) = cos(360 - x)
=> ( x + 45) = 360 - x
=> 2x = 315
x = 315/2 = 157.5°
c) sin(x-30)=1/2cosx
=>â3/2 sin x - 1/2 cos x = 1/2 cos x
=> â3 sin x = 2 cos x
divide by cos x
tan x = 2/â3
x = 49.11° and 229.11°
d) 3sin(x+10 ) = 4cos(x-10)
3(sin x cos 10 + cos x sin(10)) = 4( cos x cos 10 + sin x sin 10)
=> 2.9544 sin x + 0.5209 cos x = 3.939 cos x + 0.695 sin x
=>2.2594 sin x = 3.4181 cos x
=> tan x = 1.5128
x = 56.5349° and 236.5349°