May i know how to solve this qn: 3sinxcos40degrees = 2+ cosxsin40degrees?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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 You need to transform the following products into sums such that:

sin x*cos 40 = (1/2)*[sin(x - 40) + sin (x + 40)]

3sin x*cos 40 = (3/2)*[sin(x - 40) + sin (x + 40)]

sin 40*cos x = (1/2)*[sin(40 - x) + sin (40 + x)]

sin(40 - x)= sin(-(x - 40)) = -sin(x - 40)

Write the equation such that:

(3/2)*[sin(x - 40) + sin (x + 40)] = 2 + -sin(x - 40)/2 + sin(x + 40)/2

Moving all terms to one side yields:

(3/2)*sin(x - 40) + (3/2)* sin (x + 40) - 2 + sin(x - 40)/2 - sin(x + 40)/2 = 0

Adding like terms yields:

`(4/2)*sin(x - 40) + (2/2)*sin(x+ 40) - 2 = 0`

`2sin(x - 40) + sin(x+ 40) - 2 = 0`

If `x = 90^o =gt sin(x - 40) = sin(x+ 40)`

`2sin(x - 40) + sin(x- 40) - 2 = 0 =gt 3sin(x - 40) = 2 `

`sin(x - 40) = 2/3 =gt x - 40 = sin^-1 (2/3) + n*pi`

`x = sin^-1 (2/3) + n*pi + 40`

`` If `x =180^o=gt sin(x - 40) = -sin(x+ 40)`

`2sin(x - 40) + sin(x+ 40) - 2 = 0 =gt `

`2sin(x - 40)- sin(x- 40) - 2 = 0 =gt sin(x - 40) = 2, sin(x - 40)=lt 1` impossible

The arguments x - 40 and x + 40 must be either in quadrants 1 and 2, or in quadrants 3 and 4 and the solution to this equation is `x = sin^-1 (2/3) + n*pi + 40` .

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pagu | (Level 2) eNoter

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Solving for x in 3sinx cos40° = 2 + cosx sin40°

Then we divide both sides by 3cos40°

sinx  = (2/3cos40°) + cosx sin40°/(3cos40°)

Rearranging terms and simplifying we get

sinx  - cosx(1/3)tan40° - (2/3cos40°)  = 0

if we let A = (1/3)tan40° and B = (2/3cos40°), the equation is reduced to

sinx  - Acosx - B  = 0

Here we can't find any trigonometric identity that transform the equation into a single trigonometric function.

So we will apply the GUESS & SEE method, the objective is to find the value x to make sinx  - Acosx - B  equal to 0. It's quite discouraging to imagine the amount of work we need here but by carefully observing how the value of sinx  - Acosx - B changes as we try several values of x will lessen our work. Of course we need electronic calculator to accomplish these.

Let's begin guessing

if x=0°             sinx  - Acosx - B = -1.14997

if x=90°           sinx  - Acosx - B = 0.129728

Now it's clear that x is between 0° and 90°, more likely nearer to 90° than to 0°.

try x =70°       sinx  - Acosx - B = -0.026242

try x = 75°      sinx  - Acosx - B = 0.023262

Now it's obvious the x is almost halfway between 70° & 75°

try x = 72.5°   sinx  - Acosx - B = -0.000662

Therefore x = 72.5° may already be acceptable

I did use MS Excel What-If Analysis-GoalSeek function and quickly generate result for x = 72.5668962° with sinx  - Acosx - B = 0 rounded to 9th place

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