Sum problem.Calculate the sum of sin(45-x) + sin(45+x) . 

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sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You may also convert the sum into a product using the following formula, such that:

`sin a + sin b = 2 sin((a + b)/2)*cos((a - b)/2)`

Reasoning by analogy yields:

`sin(45 - x) + sin(45 + x) = 2sin((45 - x + 45 + x)/2)*cos((45 - x - 45 - x)/2)`

`sin(45 - x) + sin(45 + x) = 2sin(90/2)cos(-2x/2)`

`sin(45 - x) + sin(45 + x) = 2sin(45)cos(-x)`

Using the fact that the cosine function is even, yields:

`sin(45 - x) + sin(45 + x) = 2sin(45)cos x`

`sin(45 - x) + sin(45 + x) = 2*sqrt2/2*cos x`

Reducing duplicate factors yields:

`sin(45 - x) + sin(45 + x) = sqrt2*cos x `

Hence, evaluating the given sum yields `sin(45 - x) + sin(45 + x) = sqrt2*cos x.`

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll apply the sine function to the sum and the difference of angles 45 and x:

sin (a-b) = sin a*cos b - sin b*cos a

sin (a+b) = sin a*cos b + sin b*cos a

We'll put a = 45 and b = x.

sin (45+x) = sin 45*cos x + sin x*cos 45

sin (45+x) = sqrt2*cos x/2 + sqrt2*sin x/2 (1)

sin (45-x) = sin 45*cos x - sin x*cos 45

sin (45-x) = sqrt2*cos x/2 - sqrt2*sin x/2 (2)

Now, we'll substitute (1) and (2) in the sum to be calculated:

sin(45+x) + sin(45-x) = sqrt2*cos x/2 + sqrt2*sin x/2 + sqrt2*cos x/2 - sqrt2*sin x/2

We'll combine and eliminate like terms:

sin(45+x) + sin(45-x) = 2*sqrt2*cos x/2

We'll simplify and we'll get:

sin(45+x) + sin(45-x) = sqrt2*cosx

We can notice that the result of the sum is not zero, but sqrt2*cosx.

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