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3 years ago

Can you help me due at 10:00

Mathematics

2 answers:

Elena L [17]3 years ago

4 0

Answer:

Step-by-step explanation:

alexandr402 [8]3 years ago

3 0

3 I think . When adding terms, we look for terms that are variables.

Coefficients do not need to match.

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zubka84 [21]

Answer:

See Explanation Below

Step-by-step explanation:

Given

$(sin x - cos x)^2 = sec^2x - tan^2x - 2sinx.cos x.$

Required

Prove

To start with; we open the bracket on the left hand side

$(sin x - cos x)^2 = (sin x - cos x)(sin x - cos x)$

$(sin x - cos x)^2 = (sin x )(sin x - cos x)- (cos x)(sin x - cos x)$

$(sin x - cos x)^2 = sin^2 x -sinx cos x - sin xcos x + cos^2 x$

$(sin x - cos x)^2 = sin^2 x -2sinx cos x + cos^2 x$

Reorder

$(sin x - cos x)^2 = sin^2 x + cos^2 x - 2sinx cos x$

From trigonometry;

$sin^2x + cos^2x = 1$

So;

$(sin x - cos x)^2 = sin^2 x + cos^2 x - 2sinx cos x$

becomes

$(sin x - cos x)^2 = 1 - 2sinx cos x$

Also from trigonometry;

$sec^2x - tan^2x = 1$

So,

$(sin x - cos x)^2 = 1 - 2sinx cos x$

becomes

$(sin x - cos x)^2 = sec^2x - tan^2x - 2sinx cos x$

Proved

3 0

3 years ago

Giving all my points

Dominik [7]

The answer is $2.50 because you do 145-132.50=12.5
Then you divide by 5 because that’s the difference between students then the final answer is 2.50

4 0

3 years ago

Write the expression as either the sine, cosine, or tangent of a single angle. tan5x - tan2y / 1 + tan5x tan2y

Kryger [21]

$\bf \textit{Sum and Difference Identities}
\\\\
tan(\alpha + \beta) = \cfrac{tan(\alpha)+ tan(\beta)}{1- tan(\alpha)tan(\beta)}
\qquad
tan(\alpha - \beta) = \cfrac{tan(\alpha)- tan(\beta)}{1+ tan(\alpha)tan(\beta)}\\\\
-------------------------------\\\\
\cfrac{tan(5x)-tan(2y)}{1+tan(5x)tan(2y)}\implies tan(5x-2y)$