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.
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.
a) To find Tan x, we can use the identity Tan x = sin x / cos x.
Given sin x = 2/3, we need to find cos x. The Pythagorean identity for the sin and cos functions states that sin^2 x + cos^2 x = 1.
So, plugging in the value of sin x, we have (2/3)^2 + cos^2 x = 1.
Simplifying, we get 4/9 + cos^2 x = 1.
Subtracting 4/9 from both sides, we have cos^2 x = 5/9.
Taking the square root of both sides, we get cos x = ± √(5/9).
As x is an acute angle, cos x is positive. So, cos x = √(5/9).
Now, using Tan x = sin x / cos x, we have Tan x = (2/3) / (√(5/9)).
Simplifying, we get Tan x = 2√5 / 3√(5/9).
Rationalizing the denominator, we have Tan x = (2√5 / 3)(3 / √5).
Canceling out the square root terms, we have Tan x = 2 / 3.
Therefore, Tan x = 2/3.
b) To find cos (90-x), we use the identity cos (90-x) = sin x.
Given sin x = 2/3, we have cos (90-x) = 2/3.
c) Given 2cos(3t) = 3sin(3t), we can solve for t.
First, divide both sides of the equation by 3sin(3t):
2cos(3t) / 3sin(3t) = 1.
Using the trigonometric identity cos(3x) = 1 - 2sin^2(x), we can rewrite the left-hand side of the equation:
2(1 - 2sin^2(3t))/(3sin(3t)) = 1.
Simplifying, we have:
(2 - 4sin^2(3t))/(3sin(3t)) = 1.
Multiply both sides of the equation by 3sin(3t) to eliminate the denominator:
2 - 4sin^2(3t) = 3sin(3t).
Rearranging, we have:
4sin^2(3t) + 3sin(3t) - 2 = 0.
This is now a quadratic equation in terms of sin(3t). We can solve for sin(3t) using factoring or the quadratic formula.
Finally, if tan x = 5/8, we can find sin(90-x) using the identity sin(90-x) = cos x.
First, find cos x using the Pythagorean identity:
sin^2 x + cos^2 x = 1.
Substituting the value of tan x, we have:
(5/8)^2 + cos^2 x = 1.
Simplifying, we have:
25/64 + cos^2 x = 1.
Subtracting 25/64 from both sides, we have:
cos^2 x = 39/64.
Taking the square root of both sides, we have:
cos x = ± √(39/64).
But cos x is positive because tan x is positive and x is acute. So, cos x = √(39/64).
Therefore, sin(90-x) = cos x = √(39/64).