a) We draw a right triangle with the opposite side equal to 2 and the hypotenuse equal to 3, since sin x is opposite/hypotenuse. Then, using Pythagoras' theorem, we find that the adjacent side is √(3²-2²)=√5. Therefore, tan x=opposite/adjacent=2/√5. We rationalize the denominator by multiplying numerator and denominator by √5, and get tan x=2√5/5.
b) Using the fact that cos (90-x)=sin x, we can find cos (90-x)=2/3.
To solve 2 cos 3x = 3 sin 3x, we use the identity cos 2θ=1-2sin²θ and rewrite it as cos 3x = 1 - 2sin²(3x/2). Similarly, we use the identity sin 2θ=2sinθcosθ and rewrite it as sin 3x = 2sin(3x/2)cos(3x/2). Substituting these expressions into the original equation, we get:
2(1 - 2sin²(3x/2)) = 3(2sin(3x/2)cos(3x/2))
Simplifying, we get:
4sin²(3x/2) + 3cos²(3x/2) - 6sin(3x/2)cos(3x/2) - 2 = 0
We can factor this as:
(2sin(3x/2) - 3cos(3x/2))^2 - 5 = 0
Taking the square root of both sides, we get:
2sin(3x/2) - 3cos(3x/2) = ±√5
Dividing by cos(3x/2), we get:
2tan(3x/2) - 3 = ±√5 / cos(3x/2)
Using the identity tan(π/2 - θ) = 1/tanθ, we rewrite this as:
2cot(π/4 - 3x/2) - 3 = ±√5 / sin(3x/2)
Let's solve for the case when the right-hand side is positive (√5), and leave the other as an exercise. Then, we get:
2cot(π/4 - 3x/2) - 3 = √5 / sin(3x/2)
Using the identity cot(π/4 - θ) = tanθ - 1, we rewrite this as:
2tan(3x/2) - 3 - √5 = 0
Solving for tan(3x/2), we get:
tan(3x/2) = (3 + √5) / 2
Using the identity tan(π - θ) = -tanθ and the fact that x is acute (i.e., in the first quadrant), we can write:
tan(3x/2) = tan(π - x/2)
Therefore, we get:
3x/2 = π - x/2 + kπ (where k is an integer)
Solving for x, we get:
x = (4k-1)π/10
In the range 0° to 180°, we have two solutions: x = 36° and x = 144°.
If tan x=5/8, we can draw a right triangle with the opposite side equal to 5 and the adjacent side equal to 8. Then, using Pythagoras' theorem, we find that the hypotenuse is √(5²+8²)=√89. Therefore, sin x=opposite/hypotenuse=5/√89, and using the fact that sin(90-x)=cos x, we get cos x=8/√89. To find sin(90-x), we use the fact that sin(90-x)=cos x and get sin(90-x)=8/√89. Rounded to 3 significant figures, sin(90-x)=0.899.
Given that sin x=2/3 and x an acute angle, find without using tables or calculator:a)Tan x
b)cos (90-x)
Find the values of titre between 0° and 180° such that 2 cos 3 titre=3 sin 3 titre.
If tan x=5/8 find sin(90-x).Give your answer in 3 significant figures.
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