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3 years ago

Precalculus Homework help please.

Mathematics

1 answer:

aniked [119]3 years ago

3 0

Answer:

see explanation

Step-by-step explanation:

(5)

Using the trigonometric identity

sin²x + cos²x = 1 ⇒ cos²x = 1 - sin²x

Given

3sin²Θ + 2cos²Θ - 2

= 3sin²Θ + 2(1 - sin²Θ) - 2

= 3sin²Θ + 2 - 2sin²Θ - 2

= sin²Θ

(6)

Using the trigonometric identity

secx = $\frac{1}{cosx}$

Consider the left side

$\frac{cosx}{1-sin^2x}$

= $\frac{cosx}{cos^2x}$

= $\frac{1}{cosx}$ = secx = right side ⇒ verified

(7)

Using the trigonometric identity

1 + cot²x = csc²x

Given

sin²x. cot²x + sin²x ← factor out sin²x from each term

= sin²x(cot²x + 1)

= sin²x. csc²x

= sin²x × $\frac{1}{sin^2x}$ = 1

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Step-by-step explanation:

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2 years ago

Use Euler's formula to write 3+3i^3 in exponential form.

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Cite

edited Mar 6 '17 at 4:38

Richard Ambler

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asked Mar 6 '17 at 3:34

14wml

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Add a comment

1 Answer

No. It is not true that =cos()+sin(). Notice that

1=1≠cos()+sin(),

for example consider this at =0.

As a hint for figuring this out, notice that

+=ln(+)

then recall your rules for logarithms to get this to the form (+)ln().

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2 years ago

What is 4/9 - (-2/9)

Citrus2011 [14]

Hello!

$\frac{4}{9}-\frac{2}{9}$

First you had to used apply rule -(-b)=b

$\frac{4}{9}+\frac{2}{9}$

Since the denominators are equal to it should be combine by the fractions.

$\frac{4+2}{9}$

Then add 4+2=6

$\frac{6}{9}$

$\frac{4}{9}-\frac{-2}{9}$

Finally you had to cancel by the common factor of 3.

$=Answer--->\frac{2}{3}$

Hope this helps! And thank you for posting at here on Brainly. And have a great day! -Charlie :)

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