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3 years ago

Trig identities how to prove 1- (sin^2x)/(1+cosx)= cosx

Mathematics

1 answer:

balu736 [363]3 years ago

3 0

Both sides aren't equal because 1-sin^2x = cos^2x and cos^2x / 1+cosx can never equal cosx

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Solve for B. A=3B+7C

Archy [21]

Hey there!!

Given:

... A=3B+7C

Subtracting 7C on both sides:

... A-7C=3B

Dividing by B on both sides:

... (A-7C)/3=B

Hope my answer helps!

6 0

3 years ago

Simplify 16 - 12 ÷ 4 +3 (6-2) show working plz

ira [324]

Answer:

Step-by-step explanation:

16-12/4+3(6-2) subtract 6 and 2

16-12/4+3+4 divide 12 and 4

16-3+3+4 subtract 16 and 3

13+3+4 add all 3 numbers

13+7

4 0

2 years ago

How to work out 5-[5+8÷4-1]

rodikova [14]

PEMDAS (order of operations rules) require that we perform operations in a certain order: anything inside parentheses first, followed by any exponentiation, followed by mult. and div., finally followed by addition and subtraction.

Thus, we must evaluate 5+8÷4-1 first, as it's inside parentheses. Focusing on the division first, we get 5+8÷4-1 = 5 + 2 - 1, or 6.

Then we have 5 - [6], which comes out to -1.

6 0

3 years ago

Read 2 more answers

I dont know how to divide a line segment.

melamori03 [73]

From point A, draw a line segment at an angle to the given line, and about the same length. The exact length is not important. Set the compasses on A, and set its width to a bit less than one fifth of the length of the new line. Step the compasses along the line, marking off 5 arcs. Label the last one C. With the compasses' width set to CB, draw an arc from A just below it. With the compasses' width set to AC, draw an arc from B crossing the one drawn in step 4. This intersection is point D. Draw a line from D to B. Using the same compasses' width as used to step along AC, step the compasses from D along DB making 4 new arcs across the line. Draw lines between the corresponding points along AC and DB. Done. The lines divide the given line segment AB in to 5 congruent parts.

6 0

3 years ago

How do you find the limit?

coldgirl [10]

Answer:

2/5

Step-by-step explanation:

Hi! Whenever you find a limit, you first directly substitute x = 5 in.

$\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{5^2-6(5)+5}{5^2-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{25-30+5}{25-25}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{0}{0}}$

Hm, looks like we got 0/0 after directly substitution. 0/0 is one of indeterminate form so we have to use another method to evaluate the limit since direct substitution does not work.

For a polynomial or fractional function, to evaluate a limit with another method if direct substitution does not work, you can do by using factorization method. Simply factor the expression of both denominator and numerator then cancel the same expression.

From x²-6x+5, you can factor as (x-5)(x-1) because -5-1 = -6 which is middle term and (-5)(-1) = 5 which is the last term.

From x²-25, you can factor as (x+5)(x-5) via differences of two squares.

After factoring the expressions, we get a new Limit.

$\displaystyle \large{ \lim_{x\to 5}\frac{(x-5)(x-1)}{(x-5)(x+5)}}$

We can cancel x-5.

$\displaystyle \large{ \lim_{x\to 5}\frac{x-1}{x+5}}$

Then directly substitute x = 5 in.

$\displaystyle \large{ \lim_{x\to 5}\frac{5-1}{5+5}}\\

\displaystyle \large{ \lim_{x\to 5}\frac{4}{10}}\\

\displaystyle \large{ \lim_{x\to 5}\frac{2}{5}=\frac{2}{5}}$

Therefore, the limit value is 2/5.

L’Hopital Method

I wouldn’t recommend using this method since it’s too easy but only if you know the differentiation. You can use this method with a limit that’s evaluated to indeterminate form. Most people use this method when the limit method is too long or hard such as Trigonometric limits or Transcendental function limits.

The method is basically to differentiate both denominator and numerator, do not confuse this with quotient rules.

So from the given function:

$\displaystyle \large{ \lim_{x \to 5} \frac{x^2-6x+5}{x^2-25}}$

Differentiate numerator and denominator, apply power rules.

Differential (Power Rules)

$\displaystyle \large{y = ax^n \longrightarrow y\prime= nax^{n-1}$

Differentiation (Property of Addition/Subtraction)

$\displaystyle \large{y = f(x)+g(x) \longrightarrow y\prime = f\prime (x) + g\prime (x)}$

Hence from the expressions,

$\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2-6x+5)}{\frac{d}{dx}(x^2-25)}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{\frac{d}{dx}(x^2)-\frac{d}{dx}(6x)+\frac{d}{dx}(5)}{\frac{d}{dx}(x^2)-\frac{d}{dx}(25)}}$

Differential (Constant)

$\displaystyle \large{y = c \longrightarrow y\prime = 0 \ \ \ \ \sf{(c\ \ is \ \ a \ \ constant.)}}$

Therefore,

$\displaystyle \large{ \lim_{x \to 5} \frac{2x-6}{2x}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{2(x-3)}{2x}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{x-3}{x}}$

Now we can substitute x = 5 in.

$\displaystyle \large{ \lim_{x \to 5} \frac{5-3}{5}}\\

\displaystyle \large{ \lim_{x \to 5} \frac{2}{5}}=\frac{2}{5}$

Thus, the limit value is 2/5 same as the first method.

Notes:

If you still get an indeterminate form 0/0 as example after using l’hopital rules, you have to differentiate until you don’t get indeterminate form.

8 0

2 years ago