Sin alpha minus sin beta formula

Asked by maham237 @ 06/01/2021 in Mathematics viewed by 294 People

Find the exact value of the expressions cosine left parenthesis alpha plus beta right parenthesiscos(α+β), sine left parenthesis alpha plus beta right parenthesissin(α+β) and tangent left parenthesis alpha plus beta right parenthesistan(α+β) under the following conditions: cosine left parenthesis alpha right parenthesis equals StartFraction 12 Over 13 EndFractioncos(α)= 12 13, alphaα lies in quadrant IV, and sine left parenthesis beta right parenthesis equals StartFraction negative 6 Over 13 EndFractionsin(β)= −6 13, betaβ lies in quadrant III. what is tangent (alpha +Beta)I need to find answer to tan(alpha + Beta) when sin (alpha+Beta) =5square root 133 - 72/169 and cos(alpha+Beta) = -12square root 133 + 30/169. Cos 12/13 Quad 4; Sin -6/13 Quad 3

Answered by maham237 @ 06/01/2021

Answer:

$sin(\alpha+\beta)=\frac{-72+5\sqrt{133}}{169}$

$cos(\alpha+\beta)=\frac{-30-12\sqrt{133}}{169}$

$tan(\alpha+\beta=\frac{-10+\sqrt{133}}{18}}$

Step-by-step explanation:

If $cos(\alpha)= \frac{12}{13}$ , $\alpha$ lies in quadrant IV, and $cos^2(\alpha)+sin^2(\alpha)=1$ :

$sin^2(\alpha)=1-cos^2(\alpha)\\sin^2(\alpha)=1-(\frac{12}{13})^2sin^2(\alpha)=1-\frac{144}{1169}\\sin^2(\alpha)=\frac{25}{169}\\sin(\alpha)=\pm\frac{5}{13}$

Since $\alpha$ lies in quadrant IV, the value of $sin(\alpha)$ is negative so $sin(\alpha)=-\frac{5}{13}$ .

If $sin(\beta)=-\frac{6}{13}$ , $\beta$ lies in quadrant III, and $cos^2(\beta)+sin^2(\beta)=1$ :

$cos^2(\beta)=1-sin^2(\beta)\\cos^2(\beta)=1-(\frac{-6}{13})^2\\cos^2(\beta)=1-\frac{36}{169}\\cos^2(\beta)=\frac{133}{169}\\cos(\beta)=\pm\frac{\sqrt{133}}{13}$

$\beta$ lies in quadrant III, the value of $cos(\beta)$ is negative so $cos(\beta)=-\frac{\sqrt{133}}{13}$ .

The formula for the of two angles:

$sin(\alpha+\beta)=sin(\alpha)cos(\beta)+sin(\beta)cos(\alpha)\\sin(\alpha+\beta)=(-\frac{5}{13})(\frac{-\sqrt{133}}{13})+(-\frac{6}{13})(\frac{12}{13})\\sin(\alpha+\beta)=\frac{5\sqrt{133}}{169}-\frac{72}{169}\\sin(\alpha+\beta)=\frac{-72+5\sqrt{133}}{169}$

$cos(\alpha+\beta)=cos(\alpha)cos(\beta)-sin(\alpha)sin(\beta)\\cos(\alpha+\beta)=(\frac{12}{13})(-\frac{\sqrt{133}}{13})-(-\frac{5}{13})(-\frac{6}{13})\\cos(\alpha+\beta)=\frac{-12\sqrt{133}}{169}-\frac{30}{169}\\cos(\alpha+\beta)=\frac{-30-12\sqrt{133}}{169}$

For $tan(\alpha+\beta)$

$\theta=\alpha+\beta$

$tan(\theta)=\frac{cos(\theta)}{sin(\theta)}\\tan(\alpha+\beta)=\frac{cos(\alpha+\beta)}{sin(\alpha+\beta)} \\tan(\alpha+\beta)=\frac{\frac{-72+5\sqrt{133}}{169}}{\frac{-30-12\sqrt{133}}{169}} \\tan(\alpha+\beta)=\frac{(-72+5\sqrt{133})169}{(-30-12\sqrt{133})169}$

$tan(\alpha+\beta)=\frac{-72+5\sqrt{133}}{-30-12\sqrt{133}}(\frac{-30+12\sqrt{133}}{-30+12\sqrt{133}})\\tan(\alpha+\beta)=\frac{2160-150\sqrt{133}-864\sqrt{133}+60(133)}{900-144(133)}}\\tan(\alpha+\beta)=\frac{2160-1014\sqrt{133}+7980}{900-19152}}\\tan(\alpha+\beta)=\frac{10140-1014\sqrt{133}}{-18252}}\\tan(\alpha+\beta)=\frac{1014(10-\sqrt{133})}{-18252}}\\tan(\alpha+\beta)=\frac{-(10-\sqrt{133})}{18}}\\tan(\alpha+\beta)=\frac{-10+\sqrt{133}}{18}}$

Sin alpha minus sin beta formula

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