Evaluate The Following: Cos 75° Cos 15° Sin 2 Solve

Evaluate The Followinga Cos75o Cos15ob Sin Sin2 Solve The Fol

Evaluate the following: a. cos75°− cos15° b. sin − sin 2. Solve the following equations and write the general form of the solutions. a. sin4 θ + cos2 θ = 0 b. cos6 θ − cos4 θ = . Verify the following identity: = − cot(6 θ ) 4. Simplify the following: a. cos65°− sin 25° b. sin215° + sin275° c. csc( − θ ) 5. Solve the following right triangle. Round lengths to two decimal places and angles to one tenth of a degree - Calculator required: 8 A B 5 a. A = b. B = c. a = 6. From a spot 25 feet from the base of a building, the angle of elevation to the top of the building is 60°. Find the height of the building.

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Evaluate The Followinga Cos75o Cos15ob Sin Sin2 Solve The Fol

This comprehensive analysis addresses multiple trigonometric problems, involving evaluation, identities, equation solving, simplification, and application to real-world scenarios such as right triangles and height determination. The solutions utilize fundamental trigonometric identities, algebraic manipulations, and geometric principles to provide clear, step-by-step resolutions to each problem.

1. Evaluation of Trigonometric Expressions

a. cos75° − cos15°: To evaluate this expression, we employ the cosine difference identity:

• Cosine difference identity: cos A − cos B = –2 sin[(A + B)/2] sin[(A − B)/2]

Applying the values:

• A = 75°, B = 15°
• cos75° − cos15° = –2 sin( (75°+15°)/2 ) × sin( (75°−15°)/2 )

= –2 sin(45°) × sin(30°)

Using known values:

• sin(45°) = √2/2 ≈ 0.7071
• sin(30°) = 1/2 = 0.5

Therefore:

• cos75° − cos15° = –2 × 0.7071 × 0.5 ≈ –2 × 0.35355 ≈ –0.7071

b. sin − sin 2: The expression is somewhat ambiguous, but assuming it involves the basic sine function:

If interpreted as sin θ − sin 2θ, then the difference of sines identity applies:

• Difference of sines identity: sin A − sin B = 2 cos[(A + B)/2] × sin[(A − B)/2]

Applying A = θ, B = 2θ:

• sin θ − sin 2θ = 2 cos[(θ + 2θ)/2] × sin[(θ − 2θ)/2] = 2 cos(3θ/2) × sin(−θ/2)

Since sin(−x) = –sin x:

• = –2 cos(3θ/2) × sin(θ/2)

This expression simplifies to a product involving cosine and sine functions, which can be evaluated further if θ values are specified.

2. Solving Trigonometric Equations

a. sin⁴θ + cos²θ = 0

Using the Pythagorean identity: sin²θ + cos²θ = 1, we rewrite sin⁴θ as (sin²θ)²:

• Let x = sin²θ; then the equation becomes:

x² + cos²θ = 0

But cos²θ = 1 - sin²θ = 1 − x:

• x² + (1 − x) = 0

Simplify:

• x² − x + 1 = 0

Solve this quadratic for x:

• Discriminant D = (−1)² − 4 × 1 × 1 = 1 − 4 = –3 < 0

Since D < 0, there are no real solutions; thus, the original equation has no real solutions.

b. cos⁶θ − cos⁴θ = 0

Factor out cos⁴θ:

• cos⁴θ(cos²θ − 1) = 0

Set each factor equal to zero:

• cos⁴θ = 0 → cos θ = 0
• cos²θ − 1 = 0 → cos²θ = 1 → cos θ = ±1

Solutions:

• cos θ = 0 ⇒ θ = 90°, 270° + 360°n
• cos θ = 1 ⇒ θ = 0°, 360°n
• cos θ = −1 ⇒ θ = 180°, 540°n

Expressed generally, the solutions are:

• θ = 90° + 180°n
• θ = 0° + 360°n
• θ = 180° + 360°n

3. Verifying Trigonometric Identity

Given the identity: —cot(6θ) = (some expression), suppose the identity to verify is:

cot(6θ) = cos(6θ)/sin(6θ)

Using reciprocal identities, this relates to other trigonometric functions, which can be confirmed via algebraic manipulation or unit circle values, depending on the specific expression. Since the exact expression is incomplete, a general verification approach involves expressing cotangent in terms of sine and cosine and simplifying accordingly.

4. Simplification of Trigonometric Expressions

a. cos65°− sin25°

Recall that sin(90°−x) = cos x, so sin25° = cos(65°). Therefore:

cos65° − sin25° = cos65° − cos65° = 0

b. sin215° + sin275°

Use the sum-to-product identity:

• sin A + sin B = 2 sin[(A + B)/2] × cos[(A − B)/2]

Apply A=215°, B=275°:

• sin215° + sin275° = 2 sin( (215°+275°)/2 ) × cos( (215°−275°)/2 )

= 2 sin(490°/2) × cos(−60°)

sin(245°) = sin(180° + 65°) = -sin 65°

cos(−60°) = cos 60° = 0.5

Therefore:

• 2 × (−sin 65°) × 0.5 = –sin 65°

Numerically, sin 65° ≈ 0.9063:

• −0.9063

c. csc(−θ)

Recall csc(−θ) = 1/sin(−θ) = −1/sin θ, since sin(−θ) = −sin θ:

Thus, csc(−θ) = −csc θ

5. Solving a Right Triangle

Given sides a=6, and two unknown angles A and B, with the side a opposite angle A. Assume the question provides other angles or sides but is incomplete; here, the general approach is:

• Use the Law of Sines or Law of Cosines, depending on available data, to find missing angles and sides.

Suppose the goal is to find missing sides or angles with given data. For example, if angle A is known, and side a=6, and side b or angle B is known, calculations proceed accordingly, utilizing inverse trigonometric functions to find angles and the Law of Sines for sides.

Due to incomplete data, a general method involves:

• Calculating unknown angles using inverse sine or cosine functions.
• Applying the Law of Sines: a / sin A = b / sin B = c / sin C

All calculations should be rounded to one decimal place for angles and two decimal places for lengths.

6. Application: Height of a Building

A person stands 25 feet from the building, and the angle of elevation to the top is 60°. Using right triangle trigonometry:

• tan θ = opposite / adjacent

Therefore:

• tan 60° = height of building (h) / 25 ft

Since tan 60° = √3 ≈ 1.732, then:

• h = 25 × 1.732 ≈ 43.3 feet

Thus, the height of the building is approximately 43.3 feet.

This illustrates practical application of tangent in real-world measurement scenarios.

Conclusion

This comprehensive analysis synthesizes evaluation, algebraic identities, equation solving, and application of trigonometry in context. Utilizing fundamental identities like difference formulas, double-angle identities, and the Law of Sines and Cosines enables precise solutions to diverse problems. These techniques are essential tools in mathematical problem-solving and real-world applications such as surveying and architecture.

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