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mathpretrighelp
12-11-2005, 11:51 AM
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

Unco
12-11-2005, 01:03 PM
G'day,

Your work is outstanding.

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

Let's break the formula up.

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

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

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

and

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

so
\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:

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

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

\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 48\sqrt{10} - 21.

soroban
12-11-2005, 01:21 PM
Hello, mathpretrighelp!

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

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

This is what i know by solving the above problems:

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

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

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

plugging in the values from above, i got: \;\frac{-(48\sqrt{10}\,+\,21)}{175} . . . . correct!
To verify: \;\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 \sin(\alpha+\beta), i used: \:\sin(\alpha+\beta)\:=\:\sin\alpha\cdot\cos\beta\ ,+\,\cos\alpha\cdot\sin\beta

i got: \:\frac{(2\sqrt{10}\,-\,72}{25} . . . . no
\sin\alpha\cdot\cos\beta\,+\,\cos\alpha\cdot\sin\b eta\:=\:\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 \tan(\alpha+\beta)\:=\;\frac{\tan\alpha\,+\,\tan\b eta}{1\,-\,\tan\alpha\cdot\tan\beta}

i got: \:\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: \;\tan\alpha\,=\,-\frac{7}{24},\;\tan\beta\,=\,\frac{3}{2\sqrt{10}}

\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}}\righ t)}\;=\;\frac{-\frac{7}{24} + \frac{3}{2\sqrt{10}}}{1 + \frac{21}{48\sqrt{10}}}

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

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

Multiply top and bottom by \sqrt{10}:

. . \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 . . .

mathpretrighelp
12-12-2005, 08:30 PM
you guys are quick! thank you so much for your help! :D [/list]