anotha trig proof (1 Viewer)

Sylfiphy · Feb 17, 2024

Ugh guys I literally want to kms becos I need to move up in Dr du except I can’t even prove any of the Questions on the hw

BJT how do I prove that
cos35+cos45+cos+75+cos85 = 2cos5cos20

Average Boreduser · Feb 17, 2024

Sylfiphy said:
Ugh guys I literally want to kms becos I need to move up in Dr du except I can’t even prove any of the Questions on the hw

BJT how do I prove that
cos35+cos45+cos+75+cos85 = 2cos5cos20

use the sums to product identity. factorise the 2cos5 out and again use prod. to sums. and u got ur rhs.

Sylfiphy · Feb 17, 2024

Average Boreduser said:
use the sums to product identity. factorise the 2cos5 out and again use prod. to sums. and u got ur rhs.

So what am I doing wrong here

Average Boreduser · Feb 17, 2024

Sylfiphy said:
So what am I doing wrong here

You factorised 2cos5 but u still expanded 2 inside.

Average Boreduser · Feb 17, 2024

Sylfiphy said:
Ugh guys I literally want to kms becos I need to move up in Dr du except I can’t even prove any of the Questions on the hw

BJT how do I prove that
cos35+cos45+cos+75+cos85 = 2cos5cos20

I think just you should not stress about it so much, because therein lies your problem. Once you perceive this notion of 'oh dr du is testing uni level maffs- oh I'm definately getting a mystery mark'- you'd automatically overthink, muddle up formulas, etc. You got this dude. But dr du can only help u so much. The rest is up to u and how you apply your knowledge. For your foundations to build, your mindset also has to be clear. Stress inhibits this

Average Boreduser · Feb 17, 2024

Also pls start using a pen. pencil is rlly hard to see- may result in careless errors.

Sylfiphy · Feb 17, 2024

Average Boreduser said:
You factorised 2cos5 but u still expanded 2 inside.

wait but even if I wasn't doing that my answer is 2cos5cos60cos20??

Average Boreduser · Feb 17, 2024

Sylfiphy said:
wait but even if I wasn't doing that my answer is 2cos5cos60cos20??

whats the exact value of cos60? also you've forgetten the 2 you'll also get from using the identity twice.

Sylfiphy · Feb 17, 2024

Average Boreduser said:
I think just you should not stress about it so much, because therein lies your problem. Once you perceive this notion of 'oh dr du is testing uni level maffs- oh I'm definately getting a mystery mark'- you'd automatically overthink, muddle up formulas, etc. You got this dude. But dr du can only help u so much. The rest is up to u and how you apply your knowledge. For your foundations to build, your mindset also has to be clear. Stress inhibits this

I know This isn't the end of the world but I really have to move up because they skip a lot of topics for the b class because they're trying to get us ahead and I have so many gaps in my knowledge because of that 🥲

Average Boreduser · Feb 17, 2024

Sylfiphy said:
I know This isn't the end of the world but I really have to move up because they skip a lot of topics for the b class because they're trying to get us ahead and I have so many gaps in my knowledge because of that 🥲

Wayyyy too much working on here but ig might give u an idea where ur working might have screwed up

Sylfiphy · Feb 17, 2024

Average Boreduser said:
whats the exact value of cos60? also you've forgetten the 2 you'll also get from using the identity twice.

YO WTF I SHOULDVE REALISED THIS

thanks dude

Sylfiphy · Feb 17, 2024

Average Boreduser said:
View attachment 42499
Wayyyy too much working on here but ig might give u an idea where ur working might have screwed up

not to be corny but love u bro

Luukas.2 · Feb 24, 2024

Sylfiphy said:
Ugh guys I literally want to kms becos I need to move up in Dr du except I can’t even prove any of the Questions on the hw

BJT how do I prove that
cos35+cos45+cos+75+cos85 = 2cos5cos20

The solution has been posted, but on the strategy / approach front, looking at the result gives a few clues:

first, you start with sums and end up with a product, which suggests the sums-to-products formulae
second, the terms will need to be taken in pairs, so which of the three possible pairings do you want?
- since these formulae involve half the sum and difference, they will yield:
  - 1st and 2nd // 3rd and 4th yield (35 + 45) / 2 = 40^o and (45 - 35) / 2 = 5^o // (75 + 85) / 2 = 80^o and (85 - 75) / 2 = 5^o
  - 1st and 3rd // 2nd and 4th yield 55^o and 20^o // 65^o and 20^o
  - 1st and 4th // 2nd and 3rd yield 60^o and 25^o // 60^o and 15^o
- noting that the goal has a 5^o term, pairing in the order written (with the 10^o differences) appears the most promising route
- this first pairing results in a new form wit angles of 40 and 80 degrees, which will further combine to 60 degrees (known values) and the 20 degrees in the question
The second pairing will also work, as it yields the desired 20 degree angle and a pair with a 10 degree difference...

$\begin{align*} \cos{35^\circ} + \cos{45^\circ} + \cos{75^\circ} + \cos{85^\circ} &= \left(\cos{75^\circ} + \cos{35^\circ}\right) + \left(\cos{85^\circ} + \cos{45^\circ}\right) \\ &= 2\cos{\frac{75^\circ + 35^\circ}{2}}\cos{\frac{75^\circ - 35^\circ}{2}} + 2\cos{\frac{85^\circ + 45^\circ}{2}}\cos{\frac{85^\circ - 45^\circ}{2}} \quad \text{as $\cos{A} + \cos{B} = 2\cos{\frac{A + B}{2}}\cos{\frac{A - B}{2}}$} \\ &= 2\cos{\frac{110^\circ}{2}}\cos{\frac{40^\circ}{2}} + 2\cos{\frac{130^\circ}{2}}\cos{\frac{40^\circ}{2}} \\ &= 2\cos{20^\circ}\left(\cos{55^\circ} + \cos{65^\circ}\right) \\ &= 2\cos{20^\circ} \times 2\cos{\frac{55^\circ + 65^\circ}{2}}\cos{\frac{55^\circ - 65^\circ}{2}} \\ &= 4\cos{20^\circ}\cos{60^\circ}\cos{\left(-5^\circ\right)} \\ &= 4\cos{20^\circ} \times \frac{1}{2} \times \cos{5^\circ} \qquad \text{as $\cos\theta$ is an even function and $\cos{60^\circ} = \frac{1}{2}$} \\ &= 2\cos{5^\circ}\cos{20^\circ} \qquad \text{as required} \end{align*}$

As will the third pairing...

$\begin{align*} \cos{35^\circ} + \cos{45^\circ} + \cos{75^\circ} + \cos{85^\circ} &= \left(\cos{85^\circ} + \cos{35^\circ}\right) + \left(\cos{75^\circ} + \cos{45^\circ}\right) \\ &= 2\cos{\frac{85^\circ + 35^\circ}{2}}\cos{\frac{85^\circ - 35^\circ}{2}} + 2\cos{\frac{75^\circ + 45^\circ}{2}}\cos{\frac{75^\circ - 45^\circ}{2}} \quad \text{as $\cos{A} + \cos{B} = 2\cos{\frac{A + B}{2}}\cos{\frac{A - B}{2}}$} \\ &= 2\cos{\frac{120^\circ}{2}}\cos{\frac{50^\circ}{2}} + 2\cos{\frac{120^\circ}{2}}\cos{\frac{30^\circ}{2}} \\ &= 2\cos{60^\circ}\left(\cos{25^\circ} + \cos{15^\circ}\right) \\ &= 2 \times \frac{1}{2} \times 2\cos{\frac{25^\circ + 15^\circ}{2}}\cos{\frac{25^\circ - 15^\circ}{2}} \qquad \text{as $\cos{60^\circ} = \frac{1}{2}$} \\ &= 2\cos{\frac{40^\circ}{2}}\cos{\frac{10^\circ}{2}} \\ &= 2\cos{5^\circ}\cos{20^\circ} \qquad \text{as required} \end{align*}$

Hence, the strategic question becomes which path is the most obvious to you and / or which is the quickest to use / write out?

anotha trig proof (1 Viewer)

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