Math year 10 5.3: trigonometric relations (1 Viewer)

jane1820 · Oct 1, 2024

i dont understand why the questions specify this, if im being honest i dont see any use for it
can someone explain?

like i just draw a triangle, choose any side i want (except the right angle side) label the angle as X (for number 5 for example) n since we knew that cosX=2/3 then the hypotenuse is 3 n i work from there literally dont see a use for the X+Y=90 degrees bit

cafsrank1 · Oct 1, 2024

sin(90-alpha)=cosalpha

cafsrank1 · Oct 1, 2024

it makes life easier

WeiWeiMan · Oct 1, 2024

jane1820 said:
i dont understand why the questions specify this, if im being honest i dont see any use for it
can someone explain? View attachment 44402
like i just draw a triangle, choose any side i want (except the right angle side) label the angle as X (for number 5 for example) n since we knew that cosX=2/3 then the hypotenuse is 3 n i work from there literally dont see a use for the X+Y=90 degrees bit

3) a+b = 90˚ actually defines wtf b is
cos(b) = cos(90˚-a) = sin(a) = 5/13
cos(a) = sqrt(1-sin^2(a)) =12/13 = sin(b)

ngl, reading all the questions, they're just defining what the angles are

jane1820 · Oct 1, 2024

WeiWeiMan said:
3) a+b = 90˚ actually defines wtf b is
cos(b) = cos(90˚-a) = sin(a) = 5/13
cos(a) = sqrt(1-sin^2(a)) =12/13 = sin(b)

ngl, reading all the questions, they're just defining what the angles are

Yh but i solved the question without referring to that bit if ykwim?
u can see that i solved it (i usually just draft on the textbook before writing the final answer) but i didnt refer to that bit
am i doing the question right through a wrong approach?
like

WeiWeiMan · Oct 1, 2024

jane1820 said:
Yh but i solved the question without referring to that bit if ykwim?
u can see that i solved it (i usually just draft on the textbook before writing the final answer) but i didnt refer to that bit
am i doing the question right through a wrong approach?
like View attachment 44403

yeah
if alpha and beta were defined in any other way, your approach wouldn't work. the a+b=90˚ defined them as those other angles in a right angled triangle.

Luukas.2 · Oct 1, 2024

jane1820 said:
i dont understand why the questions specify this, if im being honest i dont see any use for it
can someone explain? View attachment 44402
like i just draw a triangle, choose any side i want (except the right angle side) label the angle as X (for number 5 for example) n since we knew that cosX=2/3 then the hypotenuse is 3 n i work from there literally dont see a use for the X+Y=90 degrees bit

I am guessing that you are just starting to deal with angles of any magnitude, and thus that trig functions aren't always positive.

In each case, the noted information is meant to tell you that the two angles are both acute and complementary... you need the (highlighted) complementary part to relate results from angle X to angle Y. In other words, knowing that

$\alpha + \beta = 90^\circ\ \text{ means we know that }\ \sin{\beta}=\sin{\left(90^\circ - \alpha\right)} = \cos{\alpha}\ \text{ and that }\ \cos{\beta}=\cos{\left(90^\circ - \alpha\right)} = \sin{\alpha}.$

But, you also need the acute part (or equivalent information) to deduce whether the trig functions are all positive. If both angles aren't acute then there may be two possible answers for some parts as non-acute angles can have trig functions that are negative. In other words,

$\text{to find } \sin{\alpha}\ \text{ from }\ \cos{\alpha}\ \text{ and a triangle, you need to know either that } \alpha\ \text{ is acute or in which quadrant }\ \alpha\ \text{ is located.}$

Your answers are assuming the acute part, which technically isn't justified.

$\text{For example, if } \sin{\alpha}=\frac{3}{5}\ \text{ and we know that }\ \alpha + \beta = 90^\circ,\ \text{ then we can deduce that }\ \cos{\beta}=\sin{\alpha}=\frac{3}{5}.$

$\text{However, even knowing that }\ \cos{\alpha}=\sin{\beta}\ \text{ isn't enough to know the value of these (as they could be positive or negative).}$

$\text{But, if we are given that }\ \alpha\ \text{ is an obtuse angle, i.e. }\ 90^\circ \le \alpha \le 180^\circ,\ \text{ then we can deduce that }\ \cos{\alpha} = -\frac{4}{5}$

$\text{From which we can state that }\ \sin{\beta} = -\frac{4}{5},\ \text{ which is consistent with the domain of }\ \beta\ \text{ being }\ -90^\circ\le \beta \le 0^\circ.$

$\text{If we are given that }\ \alpha + \beta = 90^\circ\ \text{ and that }\ \sin{\alpha}=\frac{3}{5},\ \text{ but no further information, then we can conclude that }\ \cos{\beta}=\frac{3}{5},$

$\text{ but we will be unable to determine whether }\ \cos{\alpha}=\sin{\beta}=+\frac{4}{5}\ \text{ or }\ \cos{\alpha}=\sin{\beta}=-\frac{4}{5}.$

jane1820 · Oct 2, 2024

Luukas.2 said:
I am guessing that you are just starting to deal with angles of any magnitude, and thus that trig functions aren't always positive.

In each case, the noted information is meant to tell you that the two angles are both acute and complementary... you need the (highlighted) complementary part to relate results from angle X to angle Y. In other words, knowing that

$\alpha + \beta = 90^\circ\ \text{ means we know that }\ \sin{\beta}=\sin{\left(90^\circ - \alpha\right)} = \cos{\alpha}\ \text{ and that }\ \cos{\beta}=\cos{\left(90^\circ - \alpha\right)} = \sin{\alpha}.$

But, you also need the acute part (or equivalent information) to deduce whether the trig functions are all positive. If both angles aren't acute then there may be two possible answers for some parts as non-acute angles can have trig functions that are negative. In other words,

$\text{to find } \sin{\alpha}\ \text{ from }\ \cos{\alpha}\ \text{ and a triangle, you need to know either that } \alpha\ \text{ is acute or in which quadrant }\ \alpha\ \text{ is located.}$

Your answers are assuming the acute part, which technically isn't justified.

$\text{For example, if } \sin{\alpha}=\frac{3}{5}\ \text{ and we know that }\ \alpha + \beta = 90^\circ,\ \text{ then we can deduce that }\ \cos{\beta}=\sin{\alpha}=\frac{3}{5}.$

$\text{However, even knowing that }\ \cos{\alpha}=\sin{\beta}\ \text{ isn't enough to know the value of these (as they could be positive or negative).}$

$\text{But, if we are given that }\ \alpha\ \text{ is an obtuse angle, i.e. }\ 90^\circ \le \alpha \le 180^\circ,\ \text{ then we can deduce that }\ \cos{\alpha} = -\frac{4}{5}$

$\text{From which we can state that }\ \sin{\beta} = -\frac{4}{5},\ \text{ which is consistent with the domain of }\ \beta\ \text{ being }\ -90^\circ\le \beta \le 0^\circ.$

$\text{If we are given that }\ \alpha + \beta = 90^\circ\ \text{ and that }\ \sin{\alpha}=\frac{3}{5},\ \text{ but no further information, then we can conclude that }\ \cos{\beta}=\frac{3}{5},$

$\text{ but we will be unable to determine whether }\ \cos{\alpha}=\sin{\beta}=+\frac{4}{5}\ \text{ or }\ \cos{\alpha}=\sin{\beta}=-\frac{4}{5}.$

OHHHH OK THX
n yh the next chapter was all abt the negatives n positives so i think that bit was to tell me that they ’wont’ be negative considering i wouldnt have known
anyways thank u

Math year 10 5.3: trigonometric relations (1 Viewer)

jane1820

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