WRITE THE SUBINTERVAL OF (0,PI) IN WHICH SIN (X+PI/4) IS INCREASING

     

We know that $cos(x) + sin(x)= sqrt2 sin(pi/4 + x)$. So, $$sin(pi/4 + x) = cos(pi/2 - x)$$ But $cos(pi/2 - x) = sin x$, so we must have $$sin(pi/4 + x)=sin(x).$$Now if I change $sin(x)$ to lớn $sin(pi - x)$ then the correct answer will come, i.e.$$sin(pi/4 + x)=sin(pi - x) implies pi/4 + x = pi -x implies x=3pi/8.$$ But why is $pi/4 + x=x$ not true. My book says, if $sin(x)=sin(y)$ then $ x = (-1)^ny + npi $. If I put $n=0$ then $x=y$ but this is not true in my case. Why?

I also looked on the relevant Wikipedia article but could not understand why these formulas are true for any value of $ heta$ that may be more that $pi/2$ & less than $0$.


*

*

My book says, if $sin(x)=sin(y)$ then $ x = (-1)^ny + npi $.

Bạn đang xem: Write the subinterval of (0,pi) in which sin (x+pi/4) is increasing

Yes, that means if $sin(x)=sin(y)$, then there exists some $n$ such that $ x = (-1)^ny + npi $, not that you get to lớn choose which $n$, or that the statement is true for every $n$. For example,$$sin(0)=0=sin(7pi)$$and we vì indeed have that $0=(-1)^npi+npi$ for $n=7$, but that is the only $n$ that works.

More generally, you should be aware that most functions do not have the property that $f(x_1)=f(x_2)implies x_1=x_2$. Take a look at the Wikipedia page on injective functions.


*

See, If $sin heta =0,$ then $ heta=npi, forall n in Bbb Z$. Similarly if $cos heta=0$, then $ heta=frac (2n+1)pi2, forall n in Bbb Z$.

Now, If $sin x=sin y$, then $sin x - sin y=0$ $Rightarrow$ $2cos(fracx+y2)sin(fracx-y2)=0$ $Rightarrow$ $cos(fracx+y2)=0$ or $sin(fracx-y2)=0$.

Xem thêm: Nghị Luận Về Tác Hại Của Facebook (23 Mẫu), Nghị Luận Về Hiện Tượng Nghiện Facebook (23 Mẫu)

If $cos(fracx+y2)=0$, then $frac x+y2=frac (2n+1)pi2, forall n in Bbb Z$ $Rightarrow$ $x=(2n+1)pi-y, forall n in Bbb Z$.If $sin(fracx-y2)=0$, then $frac x-y2=npi, forall n in Bbb Z$ $Rightarrow$ $x=2npi + y, forall nin Bbb Z$.

Combining these two results, we see that $x=(-1)^ny+npi, forall n in Bbb Z. $


*

Thanks for contributing an answer lớn goodsmart.com.vnematics Stack Exchange!

Please be sure khổng lồ answer the question. Provide details and share your research!

But avoid

Asking for help, clarification, or responding lớn other answers.Making statements based on opinion; back them up with references or personal experience.

Use goodsmart.com.vnJax lớn format equations. goodsmart.com.vnJax reference.

Xem thêm: 14 Cách Giảm Mỡ Toàn Thân Trong 1 Tuần ? Hiệu Quả Hơn Mong Đợi

To learn more, see our tips on writing great answers.


Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy


Not the answer you're looking for? Browse other questions tagged algebra-precalculus trigonometry angle or ask your own question.


Showing $sinfracpi13 cdot sinfrac2pi13 cdot sinfrac3pi13 cdots sinfrac6pi13 = fracsqrt1364$
How khổng lồ simplify $fracsqrt2+2(cos 20^circ+cos 25^circ)sin left(90-frac452 ight)sin 55^circ sin 57.5^circ$?
*

Site design / logo sản phẩm © 2022 Stack Exchange Inc; user contributions licensed under cc by-sa. Rev2022.7.19.42617


Your privacy

By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device và disclose information in accordance with our Cookie Policy.