Math Problem Statement

Simplify the trigonometric expression: \( \frac{1 - 4\sin(10^\circ)\sin(70^\circ)}{2\cos(10^\circ) + 2} \).

Solution

The mathematical expression in the image is:

14sin(10)sin(70)2cos(10)+2.\frac{1 - 4\sin(10^\circ) \cdot \sin(70^\circ)}{2\cos(10^\circ) + 2}.

Solution:

Step 1: Simplify sin(10)sin(70)\sin(10^\circ) \cdot \sin(70^\circ)

Using the trigonometric product-to-sum formula: sin(A)sin(B)=12[cos(AB)cos(A+B)],\sin(A) \sin(B) = \frac{1}{2} \left[\cos(A-B) - \cos(A+B)\right], we substitute A=10A = 10^\circ and B=70B = 70^\circ: sin(10)sin(70)=12[cos(1070)cos(10+70)].\sin(10^\circ) \cdot \sin(70^\circ) = \frac{1}{2} \left[\cos(10^\circ - 70^\circ) - \cos(10^\circ + 70^\circ)\right].

Simplify the angles: sin(10)sin(70)=12[cos(60)cos(80)].\sin(10^\circ) \cdot \sin(70^\circ) = \frac{1}{2} \left[\cos(-60^\circ) - \cos(80^\circ)\right].

Using cos(θ)=cos(θ)\cos(-\theta) = \cos(\theta), we get: sin(10)sin(70)=12[cos(60)cos(80)].\sin(10^\circ) \cdot \sin(70^\circ) = \frac{1}{2} \left[\cos(60^\circ) - \cos(80^\circ)\right].

Substitute cos(60)=12\cos(60^\circ) = \frac{1}{2}: sin(10)sin(70)=12[12cos(80)].\sin(10^\circ) \cdot \sin(70^\circ) = \frac{1}{2} \left[\frac{1}{2} - \cos(80^\circ)\right].

Simplify: sin(10)sin(70)=1412cos(80).\sin(10^\circ) \cdot \sin(70^\circ) = \frac{1}{4} - \frac{1}{2}\cos(80^\circ).

Step 2: Substitute back into the numerator

The numerator becomes: 14(1412cos(80)).1 - 4\left(\frac{1}{4} - \frac{1}{2}\cos(80^\circ)\right).

Simplify: 14(14)+4(12cos(80))=11+2cos(80).1 - 4\left(\frac{1}{4}\right) + 4\left(\frac{1}{2}\cos(80^\circ)\right) = 1 - 1 + 2\cos(80^\circ).

So the numerator is: 2cos(80).2\cos(80^\circ).

Step 3: Simplify the denominator

The denominator is: 2cos(10)+2=2(cos(10)+1).2\cos(10^\circ) + 2 = 2\left(\cos(10^\circ) + 1\right).

Step 4: Write the simplified expression

The simplified expression becomes: 2cos(80)2(cos(10)+1).\frac{2\cos(80^\circ)}{2\left(\cos(10^\circ) + 1\right)}.

Cancel the factor of 2: cos(80)cos(10)+1.\frac{\cos(80^\circ)}{\cos(10^\circ) + 1}.

Final Answer:

cos(80)cos(10)+1\boxed{\frac{\cos(80^\circ)}{\cos(10^\circ) + 1}}

Would you like further details or have additional questions?


Related Questions:

  1. How do product-to-sum trigonometric identities work?
  2. Why is cos(θ)=cos(θ)\cos(-\theta) = \cos(\theta)?
  3. How can cos(80)\cos(80^\circ) be expressed in terms of sin(10)\sin(10^\circ)?
  4. What are the common trigonometric simplifications for small angles like 1010^\circ or 8080^\circ?
  5. How does the numerator simplify step-by-step in this case?

Tip:

Always check if trigonometric identities (like product-to-sum or sum-to-product) can simplify complex products or sums of sine and cosine!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Trigonometry
Simplification of expressions
Product-to-sum identities

Formulas

Product-to-sum identity: \(\sin(A)\sin(B) = \frac{1}{2}[\cos(A-B) - \cos(A+B)]\)
Cosine symmetry: \(\cos(-\theta) = \cos(\theta)\)

Theorems

Symmetry properties of cosine
Simplification using trigonometric identities

Suitable Grade Level

Grades 10-12