Specialist 1/2 Question Thread!

[KiNSKi01]

I had the exact same question in a recent test! I get to the stage cos^2(x/2) - sin^2(x/2) and I don't know what to do  Which identity applies to this part??

[VanillaRice]

I had the exact same question in a recent test! I get to the stage cos^2(x/2) - sin^2(x/2) and I don't know what to do  Which identity applies to this part??

Has your teacher talked about the formula sheet which you receive in both exam 1 and 2?
If not, here's a link. Have a look under the trigonometric functions section - you should be able to find your answer there

Hope this helps

[KiNSKi01]

Yeh woops. forgot to make the connection that it was 1/2 theta and therefore I should be looking for a double angle identity 

I saw it instantly when I clicked on the link ahaha

[KiNSKi01]

I can't get the answer 3 root 2

Can someone please tell me where im going wrong. To find vector ED I have been finding vector OE and OD. To find OE I subtracted vector OA from OE and to find vector OD i multiplied vector OB by 0.5. When I subtract vector OD from OE and and proceed to find the square root of the two coefficients of 'i' and 'j' squared; I always get an answer larger than 3 root 2!

[Shadowxo]

I can't get the answer 3 root 2

Can someone please tell me where im going wrong. To find vector ED I have been finding vector OE and OD. To find OE I subtracted vector OA from OE and to find vector OD i multiplied vector OB by 0.5. When I subtract vector OD from OE and and proceed to find the square root of the two coefficients of 'i' and 'j' squared; I always get an answer larger than 3 root 2!

So OD = -i +2j
AB = AO + OB = -OA + OB = -4i -7j -2i +4j = -6i -3j
So AE = 1/3 * AB = -2i -j
OE = OA +  AE = 4i +7j -2i -j = 2i +6j
ED = EO + OD = -OE + OD = -2i -6j -i +2j = -3i -4j

So the length would be √(9+16) = 5. I think it's an error with the answer

[KiNSKi01]

Thank you Shadowxo!!

Thought I was going insane ahaha

[Guideme]

Hi everyone !

Can anyone explain question 8c to me please .

There isn't any examples for these type of questions :/


Thanks in advance !

[VanillaRice]

Hi everyone !

Can anyone explain question 8c to me please .

There isn't any examples for these type of questions :/
(Image removed from quote.)

Thanks in advance !

Have a sketch of the vectors a and b. You can see that the shortest distance from point A to the line OB can be represented by a line which is perpendicular to OB. In other words, this line between the line OB and point A is actually the component of a perpendicular to b! If you can find the magnitude of this 'line', you can find the shortest distance.

Hope this helps Please post if you're still stuck.

[Guideme]

Have a sketch of the vectors a and b. You can see that the shortest distance from point A to the line OB can be represented by a line which is perpendicular to OB. In other words, this line between the line OB and point A is actually the component of a perpendicular to b! If you can find the magnitude of this 'line', you can find the shortest distance.

Hope this helps Please post if you're still stuck.

Thank you for your reply! I have drawn the vector graph but struggling to find the equation of the perpendicular line

[KiNSKi01]

Hey I just did Guideme's question and got 12.5

Not sure if I got it right because of the double negative when you subtract the vector resolute of a in the direction b from a.
 My final line of working before calculating magnitude had 4i + j - (3i/2 - 3j/2)

[Shadowxo]

Hi everyone !

Can anyone explain question 8c to me please .

There isn't any examples for these type of questions :/
(Image removed from quote.)

Thanks in advance !

So shortest vector from A to the line OB is the vector component of a perpendicular to b.
To find the distance, you simply need to find the magnitude of this vector, the magnitude of the answer to b)

[KiNSKi01]

So shortest vector from A to the line OB is the vector component of a perpendicular to b.
To find the distance, you simply need to find the magnitude of this vector, the magnitude of the answer to b)

So if you aren't aware of how you calculate magnitude all you do is find the square root of the coeffecient of i squared + coeffecient of j sqaured

[Guideme]

Thank you shadowxo and kinski01!
So all you do is find the magnitude of the vector component -.-" .
I did all the hard work but didn't realise that it was just the magnitude of the last question lol.
Thank guys haha

[KiNSKi01]

Can someone help me with this question?

I took t out as a common factor then divided both sides by (2cos2pi + 2sin2pi) then I get stuck   

[Shadowxo]

Can someone help me with this question?

I took t out as a common factor then divided both sides by (2cos2pi + 2sin2pi) then I get stuck   

In this case, you can't take out t as it's inside the function (although they could have been clearer). If you could, however, you should know that cos(2π)=1 and sin(2π)=0

I think this is a harder question than you'd normally expect.