Mathematics, 10.03.2020 00:34 amusgrave2775

Hello! I need to verify this identity. Can you explain with all the steps? Thanks.

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Answer from: gin9

Both sides = $\frac{(sin\alpha+cos\alpha)}{(cos\alpha-sin\alpha)}$

Step-by-step explanation:

Let us revise some identity

sin(∝ + a) = sin(∝) cos(a) + cos(∝) sin(a)

cos(∝ + a) = cos(∝) cos(a) - sin(∝) sin(a)

tan ∝ = $\frac{sin(\alpha)}{cos(\alpha)}$

Let us simplify each side

Left hand side

∵ $tan(\alpha+\frac{\pi }{4})=\frac{sin(\alpha+\frac{\pi }{4})}{cos(\alpha+\frac{\pi}{4})}$ ⇒ (1)

∵ $sin(\alpha+\frac{\pi}{4})=sin(\alpha )cos(\frac{\pi}{4})+cos(\alpha)sin(\frac{\pi}{4})$ ⇒ (2)

∵ $sin(\frac{\pi}{4})=\frac{\sqrt{2}}{2}$ and $cos(\frac{\pi}{4})=\frac{\sqrt{2}}{2}$

- Substitute them in (2)

∴ $sin(\alpha+\frac{\pi}{4})=sin(\alpha )(\frac{\sqrt{2}}{2})+cos(\alpha)(\frac{\sqrt{2} }{2})$

∴ $sin(\alpha+\frac{\pi}{4})=(\frac{\sqrt{2}}{2})sin\alpha +(\frac{\sqrt{2} }{2})cos\alpha$

- Take $\frac{\sqrt{2}}{2}$ as a common factor

∴ $sin(\alpha+\frac{\pi}{4})=(\frac{\sqrt{2}}{2})(sin\alpha +cos\alpha)$ ⇒ (3)

∵ $cos(\alpha+\frac{\pi}{4})=cos(\alpha )cos(\frac{\pi}{4})-sin(\alpha)sin(\frac{\pi}{4})$ ⇒ (4)

∵ $cos(\frac{\pi}{4})=\frac{\sqrt{2}}{2}$ and $sin(\frac{\pi}{4})=\frac{\sqrt{2}}{2}$

- Substitute them in (4)

∴ $cos(\alpha+\frac{\pi}{4})=cos(\alpha )(\frac{\sqrt{2}}{2})-sin(\alpha)(\frac{\sqrt{2}}{2})$

∴ $cos(\alpha+\frac{\pi}{4})=(\frac{\sqrt{2}}{2})cos\alpha-(\frac{\sqrt{2} }{2})sin\alpha$

- Take $\frac{\sqrt{2}}{2}$ as a common factor

∴ $cos(\alpha+\frac{\pi}{4})=(\frac{\sqrt{2}}{2})(cos\alpha-sin\alpha)$ ⇒ (5)

Substitute (3) and (5) in (1)

∴ $tan(\alpha+\frac{\pi }{4})=\frac{\frac{\sqrt{2}}{2}(sin\alpha+cos\alpha)}{\frac{\sqrt{2}}{2}(cos\alpha-sin\alpha)}$

- Simplify it by divide up and down by $\frac{\sqrt{2}}{2}$

∴ $tan(\alpha+\frac{\pi }{4})=\frac{(sin\alpha+cos\alpha)}{(cos\alpha-sin\alpha)}$

∴ Left hand side = $\frac{(sin\alpha+cos\alpha)}{(cos\alpha-sin\alpha)}$

Right hand side

Right hand side = $\frac{cos(\alpha-\frac{5}{4}\pi )}{cos(\frac{3}{4}\pi-\alpha)}$ ⇒ (1)

∵ $cos(\alpha-\frac{5}{4}\pi)=cos(\alpha )cos(\frac{5}{4}\pi )+sin(\alpha)sin(\frac{5}{4}\pi)$ ⇒ (2)

∵ $cos(\frac{5}{4}\pi )=-\frac{\sqrt{2}}{2}$ and $sin(\frac{5}{4}\pi)=-\frac{\sqrt{2}}{2}$

- Substitute them in (2)

∴ $cos(\alpha-\frac{5}{4}\pi)=cos(\alpha )(-\frac{\sqrt{2}}{2})+sin(\alpha)(-\frac{\sqrt{2}}{2})$

∴ $cos(\alpha-\frac{5}{4}\pi )=(-\frac{\sqrt{2}}{2})cos\alpha+(-\frac{\sqrt{2} }{2})sin\alpha$

- Take $-\frac{\sqrt{2}}{2}$ as a common factor

∴ $cos(\alpha-\frac{5}{4}\pi)=(-\frac{\sqrt{2}}{2})(cos\alpha+sin\alpha)$ ⇒ (3)

∵ $cos(\frac{3}{4}\pi-\alpha)=cos(\frac{3}{4}\pi)cos(\alpha)+sin(\frac{3}{4}\pi)sin(\alpha)$ ⇒ (4)

∵ $cos(\frac{3}{4}\pi )=-\frac{\sqrt{2}}{2}$ and $sin(\frac{3}{4}\pi)=\frac{\sqrt{2}}{2}$

- Substitute them in (4)

∴ $cos(\frac{3}{4}\pi-\alpha)=(-\frac{\sqrt{2}}{2})cos(\alpha)+(\frac{\sqrt{2}}{2})sin(\alpha)$

- Take $-\frac{\sqrt{2}}{2}$ as a common factor

∴ $cos(\frac{3}{4}\pi-\alpha)=(-\frac{\sqrt{2}}{2})(cos\alpha-sin\alpha)$ ⇒ (5)

Substitute (3) and (5) in (1)

∴ $\frac{cos(\alpha-\frac{5}{4}\pi )}{cos(\frac{3}{4}\pi-\alpha)}$ = $\frac{-\frac{\sqrt{2}}{2}(cos\alpha+sin\alpha)}{-\frac{\sqrt{2}}{2}(cos\alpha-sin\alpha)}$

- Simplify it by divide up and down by $-\frac{\sqrt{2}}{2}$

∴ $\frac{cos(\alpha-\frac{5}{4}\pi )}{cos(\frac{3}{4}\pi-\alpha)}$ = $\frac{(cos\alpha+sin\alpha)}{(cos\alpha-sin\alpha)}$

∴ Right hand side = $\frac{(cos\alpha+sin\alpha)}{(cos\alpha-sin\alpha)}$

∵ cos∝ + sin∝ = sin∝ + cos∝

∴ Left hand side = Right hand side

∴ The identity is verified

Answer from: Quest

the terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly. a recursive formula for a geometric sequence with common ratio r is given by an=ran−1 a n = r a n − 1 for n≥2 n ≥ 2 .

step-by-step explanation:

Answer from: Quest

if gloria has "carry over" minutes, (that is, any minutes paid for but not used in a month can be applied to her payment for minutes in the next month,) her total usage for the three months is: total minutes = 320 + 243 + 489 = 1052 minutes number of 200 minute blocks = 1052/200 = 5 r52 (that is 5 with a remainder of 52) rounding up the number of 200 minute blocks gives her usage for the three months as 6 blocks. the cost for six blocks is 6 x $25 = $150 she also has 200 - 52 or 148 carry over minutes if gloria pays each month and unused minutes are lost, applying the rounding up procedure to each month gives: first month usage of 320 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 second month usage of 243 minutes rounds up to 2 blocks of 200 minutes for a payment of 2 x $25 = $50 third month usage of 489 minutes rounds up to 3 blocks of 200 minutes for a payment of 3 x $25 = $75 total payments for the three months are $50 + $50 + $75 = $175 get the plan with carry over minutes!

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Hello! I need to verify this identity. Can you explain with all the steps? Thanks.
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