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3 years ago

What is the range of the following function ?

Mathematics

1 answer:

vesna_86 [32]3 years ago

8 0

Answer:

All real numbers

Step-by-step explanation:

The range is whatever the output can be. In this case, the output is y. Because the line is going through all numbers vertically, and arrows are going up and down, we can say that the line is going through all y values. Therefore, the range is all real numbers

Read more about translation at:

brainly.com/question/11468584

4 0

3 years ago

Please help, trig is a real problem for me

Andreyy89

Answer:

2. cotФ = $-\frac{4}{3}$

3. sec²Ф - tan²Ф = 1

4. $\frac{sin^{2}\alpha+cos^{2}\alpha}{tan^{2}\alpha+1}$ = cos²∝

5. The value of sin x is $\frac{1}{2}$

6. cos 15° = $\frac{\sqrt{6}+\sqrt{2}}{4}$

7. tan(α + β) = $-\frac{63}{16}$

8. sin(π - Ф) = sin Ф

9. cos 2Ф = $-\frac{1}{2}$

10. sin 2Ф = 0.96

Step-by-step explanation:

∵ cos Ф = $-\frac{4}{5}$

∵ 90° < Ф < 180°

- That means Ф lies on the 2nd quadrant

∴ sin Ф is a positive value

∵ sin²Ф + cos²Ф = 1

∴ sin²Ф + ( $-\frac{4}{5}$ )² = 1

∴ sin²Ф + $\frac{16}{25}$ = 1

- Subtract from both sides $\frac{16}{25}$

∴ sin²Ф = $\frac{9}{25}$

- Take √ for both sides

∴ sinФ = $\frac{3}{5}$

∵ cotФ = cosФ ÷ sinФ

∴ cotФ = $-\frac{4}{5}$ ÷ $\frac{3}{5}$

∴ cotФ = $-\frac{4}{3}$

The expression is sec²Ф - tan²Ф

∵ tan²Ф = sec²Ф - 1

- Substitute tan²Ф by the right hand side in the expression

∴ sec²Ф - tan²Ф = sec²Ф - (sec²Ф - 1)

∴ sec²Ф - tan²Ф = sec²Ф - sec²Ф + 1

- Simplify the right hand side

∴ sec²Ф - tan²Ф = 1

The expression is $\frac{sin^{2}\alpha+cos^{2}\alpha}{tan^{2}\alpha+1}$

∵ The numerator is sin²∝ + cos²∝

∵ sin²∝ + cos²∝ = 1

∴ The numerator = 1

∵ The denominator is tan²∝ + 1

∵ tan²∝ = $\frac{sin^{2}\alpha}{cos^{2}\alpha}$

∴ tan²∝ + 1 = $\frac{sin^{2}\alpha}{cos^{2}\alpha}$ + 1

- Change 1 to fraction $\frac{cos^{2}\alpha}{cos^{2}\alpha}$

∴ tan²∝ + 1 = $\frac{sin^{2}\alpha}{cos^{2}\alpha}$ + $\frac{cos^{2}\alpha}{cos^{2}\alpha}$

- Add the two fractions

∴ tan²∝ + 1 = $\frac{sin^{2}\alpha+cos^{2}\alpha }{cos^{2}\alpha}$

∵ sin²∝ + cos²∝ = 1

∴ tan²∝ + 1 = $\frac{1}{cos^{2}\alpha}$

∴ The denominator = $\frac{1}{cos^{2}\alpha}$

∴ $\frac{sin^{2}\alpha+cos^{2}\alpha}{tan^{2}\alpha+1}$ = $\frac{1}{\frac{1}{cos^{2}\alpha}}$

- Remember denominator the denominator will be a numerator

∴ $\frac{sin^{2}\alpha+cos^{2}\alpha}{tan^{2}\alpha+1}$ = cos²∝

∵ tan x cos x = $\frac{1}{2}$

∵ tan x = $\frac{sin(x)}{cos(x)}$

∴ $\frac{sin(x)}{cos(x)}$ × cos x = $\frac{1}{2}$

- Simplify it by canceling cos x up with cos x down

∴ sin x = $\frac{1}{2}$

∴ The value of sin x is $\frac{1}{2}$

∵ cos(Ф - ∝) = cosФ cos∝ + sinФ sin∝

∵ 45 - 30 = 15

∴ cos 15° = cos(45 - 30)°

- Use the rule above to find the exact value

∵ cos(45 - 30)° = cos 45° cos 30° + sin 45° sin 30°

∵ cos 45° = $\frac{\sqrt{2}}{2}$ and sin 45° =

∵ cos 30° = $\frac{\sqrt{3}}{2}$ and sin 30° = $\frac{1}{2}$

∴ cos(45 - 30)° = $\frac{\sqrt{2}}{2}$ × $\frac{\sqrt{3}}{2}$ +

∴ cos(45 - 30)° = $\frac{\sqrt{6}}{4}$ + $\frac{\sqrt{2}}{4}$ = $\frac{\sqrt{6}+\sqrt{2}}{4}$

∴ cos 15° = $\frac{\sqrt{6}+\sqrt{2}}{4}$

∵ tan(α + β) = $\frac{tan\alpha+tan\beta}{1-tan\alpha.tan\beta}$

∵ cos α = $\frac{5}{13}$ and 0° < α < 90°

- That means α is in the 1st quadrant, then all trigonometry

ratios are positive

∵ sin²α + cos²α = 1

∴ sin²α + ( $\frac{5}{13}$ )² = 1

∴ sin²α + $\frac{25}{169}$ = 1

- Subtract $\frac{25}{169}$ from both sides

∴ sin²α = $\frac{144}{169}$

- Take √ for both sides

∴ sin α = $\frac{12}{13}$

∵ tan α = sin α ÷ cos α

∴ tan α = $\frac{12}{13}$ ÷ $\frac{5}{13}$

∴ tan α = $\frac{12}{5}$

∵ sin β = $\frac{3}{5}$ and 0° < β < 90°

∵ sin²β + cos²β = 1

∴ ( $\frac{3}{5}$ )² + cos²β = 1

∴ $\frac{9}{25}$ + cos²β = 1

- Subtract $\frac{9}{25}$ from both sides

∴ cos²β = $\frac{16}{25}$

- Take √ for both sides

∴ cos β = $\frac{4}{5}$

∵ tan β = sin β ÷ cos β

∴ tan β = $\frac{3}{5}$ ÷ $\frac{4}{5}$

∴ tan β = $\frac{3}{4}$

- Substitute the values of tan α and tan β in the tan (α + β)

∵ tan(α + β) = $\frac{tan\alpha+tan\beta}{1-tan\alpha.tan\beta}$

∴ tan(α + β) = $\frac{\frac{12}{5}+\frac{3}{4}}{1-(\frac{12}{5})(\frac{3}{4})}$

∴ tan(α + β) = $-\frac{63}{16}$

∵ sin(α - β) = sin α cos β - cos α sin β

∴ sin(π - Ф) = sin π cos Ф - cos π sin Ф

∵ sin π = 0 and cos π = -1

∴ sin(π - Ф) = (0) × cos Ф - (-1) × sin Ф

∴ sin(π - Ф) = 0 + sin Ф

∴ sin(π - Ф) = sin Ф

∵ cos 2Ф = 2 cos²Ф - 1

∵ cos Ф = $\frac{1}{2}$

∴ cos 2Ф = 2 ( $\frac{1}{2}$ )² - 1

∴ cos 2Ф = 2 × $\frac{1}{4}$ - 1

∴ cos 2Ф = $\frac{1}{2}$ - 1

∴ cos 2Ф = $-\frac{1}{2}$

10.

∵ sin 2Ф = 2 sinФ cosФ

∵ cosФ = 0.6 and 0° < Ф < 90°

- That means Ф is in the 1st quadrant and all its trigonometry

ratios are positive

∵ sin²Ф + cos²Ф = 1

∴ sin²Ф + (0.6)² = 1

∴ sin²Ф + 0.36 = 1

- Subtract 0.36 from both sides

∴ sin²Ф = 0.64

- Take √ for both sides

∴ sinФ = 0.8

- Substitute the values of sinФ and cosФ in the rule above

∴ sin 2Ф = 2(0.8)(0.6)

∴ sin 2Ф = 0.96

4 0

4 years ago

What is the product in simplest form? State any restrictions on the variable. Please show your work.

Zina [86]

I hope this helped. I did everything in one step but u can definitely show more work

4 0

3 years ago

Read 2 more answers