For the first example you showed, couldn’t I add cos(theta) to both sides and then factor out 2, then use the cos^2 + sin^2. If not why can’t I do that 5:00
Hi. I think you mean adding cos^2(θ) to both sides to arrive at 2sin^2(θ) + 2cos^2(θ) = 2 Unfortunately this isn't the method you are expected to use in a "show that" question. You want to start with the left hand side and finish at the right, or vice versa. AQA's mark scheme says: "Either starts with the left and finishes with the right or vice versa. Max M2 for any working that meets in the middle by trying to solve an equation". This means you would lose a mark for taking this approach.
Is sin^2(x) + cos^2(x) = 1 and extension of Pythagoras? Assume you have a right angled triangle with a hypotenuse of 1, by definition, the opposite side would be sinx and the adjacent side would be cosx. From this, if we apply Pythagoras theorem, we get the original sin^2(x) + cos^2(x) = 1 equation. I might be wrong, but it seemed like an interesting coincidence.
You could derive it in this way but it would prove that it was true for angles less than 90 degrees only. Since it is an identity it must be true for all values so we need something stronger for that.
@@1stClassMaths Could this not be due to the fact that the sin and cos functions are periodic? Meaning, you can still derive the overall identity, but you can plug in angles > 90 since the function repeats. I could be totally wrong about all of this, but it seems too coincidental to not at least have an explanation - perhaps something to do with the unit circle. Either way, I very much appreciate your active engagement in this line of questioning: it certainly helps me understand the course content better :)
Hi the periodic nature would mean you can demonstrate it for numbers from 360 to 450. So if it is true for say x = 10 the it must be true for x = 370. The idea of the unit circle will work to show for all angles though as you can draw the triangle in all quadrants.
Just learning trig identities for the upcoming further paper, for the question at 5:08 i did it differently, i added the sin^2 + cos^2 to get sin^2 + 1 = 2 - cos^2. then moved the cos^2 to the other side of the equation to get sin^2 + cos^2 + 1 = 2. as sin^2 + cos^2 = 1, 1+1 = 2. so therefore 2=2. is this correct?
Hi What you said makes sense but we should try to start with the LHS and end up with the RHS. Here you have manipulated it just like it's an equation. Instead using your approach I would write: 2sin^2(Θ) + cos^2(Θ) = sin^2(Θ) + sin^2(Θ) + cos^2(Θ) = sin^2(Θ) + 1 = [ 1 - cos^2(Θ) ] + 1 = 1 - cos^(Θ) + 1 = 2 - cos^2(Θ)
Hi. It is not ignored, we just choose not to use it in the factorisation (in fact we cannot use it as it has no sin) So the sin^4x + sin^2xcos^2x becomes sin^2x(sin^2x + cos^2x) In a simpler example consider 5x^2 + 3x + 4 You could factorise x from the first two terms only so it would be 5x^2 + 3x + 4 = x(5x + 3) + 4
bro saved me, im trying to learn 100 hours of content the night before the exam...
Glad I could help
me too
lmao same im soo finished, ima sleep now and doo chem at 4am in the morning. im too cooked
Thanks for helping me too!@@1stClassMaths
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Brilliant explanations 👌🏽
Your explanation is top-notch 👍. You're are saving me.
This video was super helpful!!! Wishing the best for you!!!! Thank you so much!!!!
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For the first example you showed, couldn’t I add cos(theta) to both sides and then factor out 2, then use the cos^2 + sin^2. If not why can’t I do that 5:00
Wait is it cos it’s an identity and I can add stuff on both sides
Hi. I think you mean adding cos^2(θ) to both sides to arrive at 2sin^2(θ) + 2cos^2(θ) = 2
Unfortunately this isn't the method you are expected to use in a "show that" question. You want to start with the left hand side and finish at the right, or vice versa.
AQA's mark scheme says: "Either starts with the left and finishes with the right or vice versa. Max M2
for any working that meets in the middle by trying to solve an equation". This means you would lose a mark for taking this approach.
Anyone here 1 hour before the exam?
yes
I’m just naturally curious about the universe. No test
I'm here 1 day before it lol
Brilliant video once again!!! You have now become my trusted go-to online tutor 🙏 thank you for all you do
Wow, thank you!
Is sin^2(x) + cos^2(x) = 1 and extension of Pythagoras? Assume you have a right angled triangle with a hypotenuse of 1, by definition, the opposite side would be sinx and the adjacent side would be cosx. From this, if we apply Pythagoras theorem, we get the original sin^2(x) + cos^2(x) = 1 equation. I might be wrong, but it seemed like an interesting coincidence.
This does make sense yes but the identity is stronger as it will work for non right angled triangles too and angles greater than 90 degrees.
@@1stClassMaths I see. I guess I was meaning that the identity could potentially be derived using logic similar to this.
You could derive it in this way but it would prove that it was true for angles less than 90 degrees only. Since it is an identity it must be true for all values so we need something stronger for that.
@@1stClassMaths Could this not be due to the fact that the sin and cos functions are periodic? Meaning, you can still derive the overall identity, but you can plug in angles > 90 since the function repeats. I could be totally wrong about all of this, but it seems too coincidental to not at least have an explanation - perhaps something to do with the unit circle. Either way, I very much appreciate your active engagement in this line of questioning: it certainly helps me understand the course content better :)
Hi the periodic nature would mean you can demonstrate it for numbers from 360 to 450. So if it is true for say x = 10 the it must be true for x = 370. The idea of the unit circle will work to show for all angles though as you can draw the triangle in all quadrants.
got some of it but the last one was to difficult . Understood alot of it though
Just learning trig identities for the upcoming further paper, for the question at 5:08 i did it differently, i added the sin^2 + cos^2 to get sin^2 + 1 = 2 - cos^2. then moved the cos^2 to the other side of the equation to get sin^2 + cos^2 + 1 = 2. as sin^2 + cos^2 = 1, 1+1 = 2. so therefore 2=2. is this correct?
Hi
What you said makes sense but we should try to start with the LHS and end up with the RHS. Here you have manipulated it just like it's an equation. Instead using your approach I would write:
2sin^2(Θ) + cos^2(Θ)
= sin^2(Θ) + sin^2(Θ) + cos^2(Θ)
= sin^2(Θ) + 1
= [ 1 - cos^2(Θ) ] + 1
= 1 - cos^(Θ) + 1
= 2 - cos^2(Θ)
Bro thank you so much for the vids they really helped today on paper 1
Glad to hear it
Don't mind but why these equations are basic
In CIE syll, they are learnt in pure maths which is also later used in further math
Because this is further maths GCSE not a-level, so the content isn't going to be as advanced as a-level.
When you factorise sin^4x + sin^2xcos^2x + cos^2x, why are you able to just ignore the cos^2x
Hi. It is not ignored, we just choose not to use it in the factorisation (in fact we cannot use it as it has no sin)
So the sin^4x + sin^2xcos^2x becomes sin^2x(sin^2x + cos^2x)
In a simpler example consider 5x^2 + 3x + 4
You could factorise x from the first two terms only so it would be 5x^2 + 3x + 4 = x(5x + 3) + 4
if this topic comes on paper 2 im finished
Frr I already flopped paper 1 💀
okay thanks
will we get these equations in the exam, do you know?
No idea - only the examiner writing them will know for sure!
You get them in the insert and you can re-arrange it
im cooked
I'm learning all topics the night before, dw you're okay
@@JanaJana-de7yb same,😂 btw how did you do on the test😅?
@@Sandy-r3m not amazing obviously but not as bad as I thought, wbu how'd you do?
u r jesus