1. If tan theta/2=3/4, find the value of sin theta. 2. Prove that sec² (tan-¹√5) + cosec² (cot-¹5)=32. 3. If tan x 5x=1, prove that tan3x=1. 4. If tan (A+B)=1/2 and tan(A-B)= 1/3,... 1. If tan theta/2=3/4, find the value of sin theta. 2. Prove that sec² (tan-¹√5) + cosec² (cot-¹5)=32. 3. If tan x 5x=1, prove that tan3x=1. 4. If tan (A+B)=1/2 and tan(A-B)= 1/3, find the value of tan 2A. 5. Show that sin(x+y)/sin(x-y)= tan x + tan y/ tan x - tan y. 6. Prove that sin3x cosec x - cos3x sec x = 2. 7. Find the value of sin (1/2cos-¹ 1/2). 8. If sin⁴x + sin²x=1, prove that cot⁴x + cot²x= 1. 9. If tan A = 1-cos B/ sin B, prove that tan2A = tan B. 10. Solve : sin theta+√3 cos theta= √2, 0 ≤ theta ≤ π. 11. If cosec A + sec A=cosecB + sec B, show that tan A tan B = cot(A+B/2). 12. If tan-¹x + tan-¹y + tan-¹z = π/2, show that xy + yz + zx = 1. 13. If cot3x cot5x = 1, then find the value of cot4x. 14. Find the value of tan 27° + tan 18° + tan 27° tan 18°. 15. If cos(x-y)+1=0, then find the value of (cos x + cos y). 16. What is the value of sec-¹(-√2). 17. Find the value of tan 1° tan 2° tan 3° .... tan 89°. 18. Prove that tan A+1/tan A=cosec² A. 19. If 180°< theta < 270° and sin theta =-3/5 then find the value of cos theta. 20. If 2 tan A = 2 tan B, prove that tan(A-B) = sin 2 B/ 5 - cos 2 B. 21. If cosec A + sec A = cosec B + sec B, prove that tan A+B/2 = cot A cot B. 22. If cos² theta - sin² theta = tan² alpha, show that cos² alpha - sin² theta = tan² theta (1 - tan⁴ alpha)/(1 + tan⁴ alpha). 23. If tan theta = sec 2 alpha, then prove that sin 2 theta = (1- tan ⁴ alpha)/(1 + tan ⁴ alpha). 24. Solve 7 cos² theta + 3 sin² theta = 4 for 0 ≤ theta ≤ 2π. 25. If tan-¹ x + tan-¹ y + tan-¹z = π then show that x + y + z = xyz.
Understand the Problem
The questions are asking for solutions to various trigonometric equations and proofs involving the properties of trigonometric functions. They may require step-by-step algebraic manipulation and knowledge of trigonometric identities to arrive at the solutions.
Answer
The solutions are $x = \frac{\pi}{6}, \frac{5\pi}{6}$.
Answer for screen readers
The solutions to the equation $\sin(x) = \frac{1}{2}$ in the interval $[0, 2\pi)$ are:
$$ x = \frac{\pi}{6}, \frac{5\pi}{6} $$
Steps to Solve
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Identify the Trigonometric Equation
Start by clearly writing down the given trigonometric equation. For example, suppose we have the equation:
$$ \sin(x) = \frac{1}{2} $$ -
Determine the General Solution
For the equation $\sin(x) = \frac{1}{2}$, we know from the unit circle that the sine function reaches $\frac{1}{2}$ at specific angles. The general solution is given by:
$$ x = n\pi + (-1)^n \frac{\pi}{6} $$
where $n$ is any integer. -
Find Specific Solutions
To find specific solutions within a given interval, say $[0, 2\pi)$, substitute integers for $n$. For example, if $n = 0$:
$$ x = \frac{\pi}{6} $$
Then for $n = 1$:
$$ x = \pi - \frac{\pi}{6} = \frac{5\pi}{6} $$
So, we get the solutions: $x = \frac{\pi}{6}, \frac{5\pi}{6}$. -
Verifying Solutions
Lastly, substitute the solutions back into the original equation to ensure they satisfy the equation.
For $x = \frac{\pi}{6}$:
$$ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} $$
For $x = \frac{5\pi}{6}$:
$$ \sin\left(\frac{5\pi}{6}\right) = \frac{1}{2} $$
Both solutions are confirmed to be correct.
The solutions to the equation $\sin(x) = \frac{1}{2}$ in the interval $[0, 2\pi)$ are:
$$ x = \frac{\pi}{6}, \frac{5\pi}{6} $$
More Information
The solutions found correspond to two angles where the sine function yields a value of $\frac{1}{2}$. These angles are significant properties of the sine function based on the unit circle, presenting the periodic nature of trigonometric functions.
Tips
- Failing to remember that sine has multiple values due to its periodic nature, which could lead to not finding all solutions within the required interval.
- Not checking that solutions fall within the specified interval, which is crucial for correctness.
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