sum and difference of cos sin and tan

mathpretrighelp

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find the exact value of the expressions cos (alpha +beta), sin (alpha+beta),tan (alpha+beta)

given cos (alpha)=24/25, alpha lies in quadrant 4 and sin (beta)= -3/7, beta lies in quadrant 3


what i am having trouble in is mainly the algebra part of it. i understand the concepts in getting the answers, but when i start dealing with radicals i get really confuse.

this is what i know by solving the above problems( i dont know if i solve the bolded ones correctly, but this may be the reason why my answers the problems look funny. the ones in bold are the ones i solved)
cos (alpha)=24/25
sin (alpha) = -7/25
sin (beta) = -3/7
cos (beta) = -2 sqrt 10/7

solving for cos (alpha +beta). i used the formula
cos (alpha +beta)= cos alpha cos beta-sin alpha sin beta
pluging in the values from above
i got
-(48 sqrt 10 + 21)/175

solving for sin (alpha+beta) i used sin
(alpha+beta)=sin alpha cos beta + cos alpha sin beta
i got
(2 sqrt 10 - 72)/25

solving for
tan (alpha+beta)=(tan alpha+ tan beta)/(1- tan alpha tan beta)
i got (-140+72 sqrt 10)/(480 -21 sqrt 10)
i know this answer is totally wrong.

any help is muchly appreciated

thanks in advance!
 
G'day,

Your work is outstanding.

mathpretrighelp said:
find the exact value of the expressions cos (alpha +beta), sin (alpha+beta),tan (alpha+beta)

given cos (alpha)=24/25, alpha lies in quadrant 4 and sin (beta)= -3/7, beta lies in quadrant 3


what i am having trouble in is mainly the algebra part of it. i understand the concepts in getting the answers, but when i start dealing with radicals i get really confuse.

this is what i know by solving the above problems( i dont know if i solve the bolded ones correctly, but this may be the reason why my answers the problems look funny. the ones in bold are the ones i solved)
cos (alpha)=24/25
sin (alpha) = -7/25
sin (beta) = -3/7
cos (beta) = -2 sqrt 10/7

solving for cos (alpha +beta). i used the formula
cos (alpha +beta)= cos alpha cos beta-sin alpha sin beta
pluging in the values from above
i got
-(48 sqrt 10 + 21)/175

solving for sin (alpha+beta) i used sin
(alpha+beta)=sin alpha cos beta + cos alpha sin beta
i got
(2 sqrt 10 - 72)/25
You may have divided the top and bottom by 7 to simplify, but haven't done so for 72 (which isn't a muliply of 7 anyway).

solving for
tan (alpha+beta)=(tan alpha+ tan beta)/(1- tan alpha tan beta)
i got (-140+72 sqrt 10)/(480 -21 sqrt 10)
i know this answer is totally wrong.
You may have become a a little bit mixed up here.

You probably had \(\displaystyle \tan{\alpha}= \frac{-7}{24}\) and \(\displaystyle \tan{\beta} = \frac{3}{2\sqrt{10}}\) (from tan=sin/cos).

Let's break the formula up.

\(\displaystyle \L \tan{(\alpha} + \tan{\beta)} = \frac{-7}{24} + \frac{3}{2\sqrt{10}}\)

The lowest common denominator is \(\displaystyle 24\sqrt{10}\):

\(\displaystyle \L = \frac{-7\sqrt{10} + 36}{24\sqrt{10}}\) <- numerator of formula

and

\(\displaystyle \L \tan{\alpha}\tan{\beta} = \frac{-7}{24} \times \frac{3}{2\sqrt{10}} = \frac{-21}{48\sqrt{10}}\)

so
\(\displaystyle \L 1 - \tan{\alpha}\tan{\beta} = 1 + \frac{21}{48\sqrt{10}} = \frac{48\sqrt{10} + 21}{48\sqrt{10}}\) <- denominator of formula

Now put these together. Multiplying the numerator by the reciprocal of the denominator:

\(\displaystyle \L \frac{-7\sqrt{10} + 36}{24\sqrt{10}} \times \frac{48\sqrt{10}}{48\sqrt{10} + 21}\)

\(\displaystyle \L 24\sqrt{10}\) cancels and we get:

\(\displaystyle \L \tan{(\alpha + \beta)} = \frac{-14\sqrt{10} + 72}{48\sqrt{10} + 21}\)

If you have the energy you can rationalise this by multiplying the top and bottom by \(\displaystyle 48\sqrt{10} - 21\).
 
Hello, mathpretrighelp!


Find the exact value of the expressions \(\displaystyle \cos(\alpha+\beta),\:\sin(\alpha + \beta), \:\tan(\alpha + \beta)\)

given \(\displaystyle \cos(\alpha)\,=\,\frac{24}{25},\,\alpha\) in Q4, and \(\displaystyle \sin(\beta)\,=-\frac{3}{7},\,\beta\) in Q3.

This is what i know by solving the above problems:

\(\displaystyle \cos(\alpha)\,=\,\frac{24}{25}\;\;\Rightarrow\;\;\sin(\alpha)\,=-\frac{7}{25}\)

\(\displaystyle \sin(\beta)\,=-\frac{3}{7}\;\;\Rightarrow\;\;\cos(\beta)\,=-\frac{2\sqrt{10}}{7}\)


Solving for \(\displaystyle \cos(\alpha +\beta)\), i used: \(\displaystyle \:\cos(\alpha+\beta)\:=\:\cos\alpha\cdot\cos\beta\,-\,\sin\alpha\cdot\sin\beta\)

plugging in the values from above, i got: \(\displaystyle \;\frac{-(48\sqrt{10}\,+\,21)}{175}\) . . . . correct!
To verify: \(\displaystyle \;\cos(\alpha + \beta)\:=\:\left(\frac{24}{25}\right)\left(-\frac{2\sqrt{10}}{7}\right)\,-\,\left(-\frac{7}{25}\right)\left(-\frac{3}{7}\right)\;=\;\frac{-48\sqrt{10}\,-\,21}{175}\)


Solving for \(\displaystyle \sin(\alpha+\beta)\), i used: \(\displaystyle \:\sin(\alpha+\beta)\:=\:\sin\alpha\cdot\cos\beta\,+\,\cos\alpha\cdot\sin\beta\)

i got: \(\displaystyle \:\frac{(2\sqrt{10}\,-\,72}{25}\) . . . . no
\(\displaystyle \sin\alpha\cdot\cos\beta\,+\,\cos\alpha\cdot\sin\beta\:=\:\left(-\frac{7}{25}\right)\left(-\frac{2\sqrt{10}}{7}\right)\,+\,\left(\frac{24}{25}\right)\left(-\frac{3}{7}\right)\:=\:\frac{14\sqrt{10}\,-\,72}{175}\)


Solving for \(\displaystyle \tan(\alpha+\beta)\:=\;\frac{\tan\alpha\,+\,\tan\beta}{1\,-\,\tan\alpha\cdot\tan\beta}\)

i got: \(\displaystyle \:\frac{-140\,+\,72\sqrt{10}}{480\,-\,21\sqr{10}}\)
- - - - -
i know this answer is totally wrong. . . . . Well, not totally
From your preliminary work, we also have: \(\displaystyle \;\tan\alpha\,=\,-\frac{7}{24},\;\tan\beta\,=\,\frac{3}{2\sqrt{10}}\)

\(\displaystyle \L\frac{\tan\alpha\,+\,\tan\beta}{1\,-\,\tan\alpha\cdot\tan\beta}\;=\;\frac{\left(-\frac{7}{24}\right)\,+\,\left(\frac{3}{2\sqrt{10}}\right)}{1\,-\,\left(-\frac{7}{24}\right)\left(\frac{3}{2\sqrt{10}}\right)}\;=\;\frac{-\frac{7}{24} + \frac{3}{2\sqrt{10}}}{1 + \frac{21}{48\sqrt{10}}}\)

Multiply top and bottom by \(\displaystyle 48\sqrt{10}:\;\;\L\frac{-14\sqrt{10}\,+\,72}{48\sqrt{10}\,+\,21}\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Multiply top and bottom by \(\displaystyle \sqrt{10}\):

. . \(\displaystyle \L\frac{\sqrt{10}}{\sqrt{10}}\cdot\frac{-14\sqrt{10}\,+\,72}{48\sqrt{10}\,+\,21}\;=\;\frac{-140\,+\,72\sqrt{10}}{480\,+\,21\sqrt{10}}\)

See? . . . Your answer was quite close . . .
 
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