Inverse functions

[Jordannn1990]

Hello im a sophomore in college and need help with a problem.

1. Simplify 2|cos(θ)| if θ = sin^-1(x/2) for some real number x.

2|cos(θ)|=?

Don't know how to do this

2.

Add and subtract as indicated. Then simplify your answer if possible. Leave answer in terms of sin(θ) and/or cos(θ). (Remember to enter trigonometric powers such as sin²(x) as (sin(x))².)

3.

Add and subtract as indicated. Then simplify your answer if possible. Leave your answer in terms of sin(θ) and/or cos(θ)

 

[HallsofIvy]

Hello im a sophomore in college and need help with a problem.

1. Simplify 2|cos(θ)| if θ = sin^-1(x/2) for some real number x.

2|cos(θ)|=?

Don't know how to do this

Do you know that \(\displaystyle sin^2(\theta)+ cos^2(\theta)= 1\)? From that and \(\displaystyle sin(\theta)= x/2\), it follows that \(\displaystyle x^2/4+ cos^2(\theta)= 1\), \(\displaystyle cos^2(\theta)= 1- x^2/4\)

2.

Add and subtract as indicated. Then simplify your answer if possible. Leave answer in terms of sin(θ) and/or cos(θ). (Remember to enter trigonometric powers such as sin²(x) as (sin(x))².)

Surely you learned long ago that to add fractions? \(\displaystyle 1/cos(\theta)+ cos(\theta)/1\) has denominators \(\displaystyle cos(\theta)\) and 1 so the "least common denominator" is \(\displaystyle cos(\theta)\).

3.

Add and subtract as indicated. Then simplify your answer if possible. Leave your answer in terms of sin(θ) and/or cos(θ)

Same thing! Why are you making no effort to do the problems your self? Do you not understand that, at base, these problems involve thing you learned in the fourth or fifth grade?

 

[Jason76]

Same thing! Why are you making no effort to do the problems your self? Do you not understand that, at base, these problems involve thing you learned in the fourth or fifth grade?

Pretty much, the adding of subtraction of fractional rational expressions as well fractonal trig stuff involves a simple concept. If more people could understand that, then they'd do better. On the surface, rational expressions and trig stuff have big numbers and look difficult, but it's a disguise.

It's kind of like what they call in war terms "a paper tiger".

For instance

\(\displaystyle \frac{2x(x + 2)^{4}}{(2x^{2} - 1)^{3}} + \frac{4x^{2}(x + 2)^{3}}{(2x^{2} - 1)^{3}} - \frac{12x^{3}(x + 2)^{4}}{(2x^{2} - 1)^{4}} \)


The LCD is \(\displaystyle (2x - 1)^{4}\)

So you make each bottom fraction equal the LCD by using multiplication (of the denominator and numerator of each fraction) by the LCD.

\(\displaystyle \frac{2x(x + 2)^{4}}{(2x^{2} - 1)^{3}} \frac{(2x^{2} - 1)}{(2x^{2} - 1)} + \frac{4x^{2}(x + 2)^{3}}{(2x^{2} - 1)^{3}} \frac{(2x^{2} - 1)}{(2x^{2} - 1)} - \frac{12x^{4}(x + 2)^{4}}{(2x^{2} - 1)^{4}} \)

\(\displaystyle \frac{2x(x + 2)^{4} (2x^{2} - 1)}{ (2x^{2} - 1)^{4}} + \frac{4x^{2}(x + 2)^{3} (2x^{2} - 1)}{ (2x^{2} - 1)^{4}} - \frac{12x^{3}(x + 2)^{4}} {(2x^{2} - 1)^{4}}\)

Now you would add and subtract the numerators for a final answer.