Math Problem Statement

sin(cos−1(square root2/2)+ tan−1(square root3))

Solution

Let's break down and solve the expression step by step:

You have: sin(cos1(22)+tan1(3))\sin \left( \cos^{-1} \left( \frac{\sqrt{2}}{2} \right) + \tan^{-1} \left( \sqrt{3} \right) \right)

Step 1: Understand the individual components

cos1(22)\cos^{-1} \left( \frac{\sqrt{2}}{2} \right)

This is the angle θ\theta such that: cos(θ)=22\cos(\theta) = \frac{\sqrt{2}}{2} We know that cos(π4)=22\cos \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2}, so: cos1(22)=π4\cos^{-1} \left( \frac{\sqrt{2}}{2} \right) = \frac{\pi}{4}

tan1(3)\tan^{-1} \left( \sqrt{3} \right)

This is the angle ϕ\phi such that: tan(ϕ)=3\tan(\phi) = \sqrt{3} We know that tan(π3)=3\tan \left( \frac{\pi}{3} \right) = \sqrt{3}, so: tan1(3)=π3\tan^{-1} \left( \sqrt{3} \right) = \frac{\pi}{3}

Step 2: Substitute the values into the expression

Now we substitute the values for the inverse trigonometric functions: sin(π4+π3)\sin \left( \frac{\pi}{4} + \frac{\pi}{3} \right)

Step 3: Simplify the angle

Let's add the angles: π4+π3=3π12+4π12=7π12\frac{\pi}{4} + \frac{\pi}{3} = \frac{3\pi}{12} + \frac{4\pi}{12} = \frac{7\pi}{12}

Step 4: Find the sine of the resulting angle

We need to calculate: sin(7π12)\sin \left( \frac{7\pi}{12} \right) We can use a sine addition formula: sin(a+b)=sin(a)cos(b)+cos(a)sin(b)\sin(a + b) = \sin(a)\cos(b) + \cos(a)\sin(b) where a=π4a = \frac{\pi}{4} and b=π3b = \frac{\pi}{3}.

Substitute the known values for sin\sin and cos\cos of these angles: sin(π4)=22,cos(π4)=22\sin \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2}, \quad \cos \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} sin(π3)=32,cos(π3)=12\sin \left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2}, \quad \cos \left( \frac{\pi}{3} \right) = \frac{1}{2}

Now, apply the sine addition formula: sin(7π12)=(22×12)+(22×32)\sin \left( \frac{7\pi}{12} \right) = \left( \frac{\sqrt{2}}{2} \times \frac{1}{2} \right) + \left( \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} \right) =24+64= \frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4} =2+64= \frac{\sqrt{2} + \sqrt{6}}{4}

Thus, the value of the given expression is: sin(cos1(22)+tan1(3))=2+64\sin \left( \cos^{-1} \left( \frac{\sqrt{2}}{2} \right) + \tan^{-1} \left( \sqrt{3} \right) \right) = \frac{\sqrt{2} + \sqrt{6}}{4}


Would you like a detailed explanation of any of the steps, or have other questions? Here are some related questions that could be useful:

  1. What are the values of cos1(12)\cos^{-1} \left( \frac{1}{2} \right) and tan1(1)\tan^{-1} (1)?
  2. How do you simplify trigonometric expressions with multiple inverse functions?
  3. Can you derive the sine addition formula and apply it to other problems?
  4. How do you calculate sine, cosine, and tangent for special angles like π6,π4,π3\frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}?
  5. What are the common trigonometric identities for solving problems involving angles?

Tip: Memorizing key values for trigonometric functions at common angles (π6,π4,π3\frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, etc.) can save you time when solving these types of problems.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Inverse Trigonometric Functions
Sine Function
Angle Addition Formula

Formulas

sin(a + b) = sin(a)cos(b) + cos(a)sin(b)
cos^(-1)(x) = angle where cos(angle) = x
tan^(-1)(x) = angle where tan(angle) = x

Theorems

Sine Addition Formula
Inverse Trigonometric Functions

Suitable Grade Level

Grades 10-12